JC2 H2 Maths Bukit Timah | Hyperbola Graphs

Did you know? A hyperbola can open left and right or up and down, and the sign of one squared term tells you which. For x²/9−y²/4=1, the two branches open horizontally, with vertices at (−3,0) and (3,0). For y²/4−x²/9=1, the branches open vertically, with vertices at (0,−2) and (0,2). The asymptote lines are the same, but the hyperbola itself is different.

This JC2 H2 Mathematics Tutor Bukit Timah guide answers a specific A-Level 9758 graphing problem: how do you recognise the branch direction, centre, vertices and asymptotes of a hyperbola, including when its centre has been translated? It starts with the answer and two clean graphs in words, then gives a complete comparison, 32 exact calculation clinics and a short review routine. The examples use hypothetical coordinate geometry, not real local properties or measured data.

The official 2027 SEAB H2 Mathematics 9758 syllabus includes the standard hyperbola graphs in Functions and Graphs §1.2, along with their symmetries, asymptotes, axes and restrictions. This page focuses on that graph-recognition skill and does not repeat the existing article about removable holes in rational functions. The immutable Clementi Mathematics small-group reference remains unchanged and provides a separate teaching-format example.

Quick answer: the positive squared term identifies the transverse direction

For positive a and b, consider two standard equations centred at (h,k). The horizontal hyperbola is (x−h)²/a²−(y−k)²/b²=1. The vertical hyperbola is (y−k)²/b²−(x−h)²/a²=1. In the first, the positive square belongs to x, so the branches open left and right; in the second, the positive square belongs to y, so the branches open up and down.

Feature	Positive x-square first	Positive y-square first
Branch direction	Left and right	Up and down
Centre	(h,k)	(h,k)
Vertices	(h−a,k) and (h+a,k)	(h,k−b) and (h,k+b)
Asymptotes	y−k=±(b/a)(x−h)	y−k=±(b/a)(x−h)
Domain	x≤h−a or x≥h+a	All real x
Range	All real y	y≤k−b or y≥k+b

First: move constants to the right and put the hyperbola into a positive-one standard form.
Second: read the centre (h,k) from the shifts, taking care with signs inside brackets.
Third: identify which squared term is positive and the associated axis of the two branches.
Fourth: locate the two vertices at distance a horizontally or b vertically from the centre.
Fifth: draw the two diagonally crossing asymptote lines through the centre using slopes ±b/a.
Sixth: sketch the branches outward from the vertices and verify the domain and range.

Worked example: a horizontal hyperbola translated from the origin

Sketch (x−1)²/9−(y+2)²/4=1. The centre is (1,−2), with a=3 and b=2. The x-square is positive, so the branches open left and right. The vertices are (1−3,−2)=(−2,−2) and (1+3,−2)=(4,−2). The asymptotes are y+2=±(2/3)(x−1).

The graph has x-domain x≤−2 or x≥4, because (x−1)²/9 must be at least one. The range includes every real y: for each chosen y, a sufficiently large positive or negative x can satisfy the equation. A quick check is the right-hand vertex (4,−2): its left-side value is (3²)/9−0=1. The centre (1,−2) is not on the curve because the left side there is zero, not one.

Worked example: a vertical hyperbola with the same centre

Now change only the order of the squares: (y+2)²/4−(x−1)²/9=1. The centre remains (1,−2), with a=3 and b=2, and the two asymptotes remain y+2=±(2/3)(x−1). But the positive y-square makes the branches vertical. The vertices become (1,−4) and (1,0), and the range becomes y≤−4 or y≥0.

This comparison demonstrates why drawing only the asymptotes cannot identify the hyperbola. Both different equations have the same pair of asymptotes. You also need to check which variable has the positive squared coefficient after normalisation and to place the vertices on the correct axis. Otherwise a neatly drawn pair of diagonals can accompany the wrong two branches.

Why the asymptotes pass through the centre

For the horizontal equation (x−h)²/a²−(y−k)²/b²=1, solve for y on the upper or lower branch: y−k=±b√[(x−h)²/a²−1]. When |x−h| becomes large, the subtractive constant one matters less relative to the squared term. The branches approach straight lines with slopes b/a and −b/a through (h,k), giving y−k=±(b/a)(x−h).

The asymptotes are not generally part of the hyperbola itself; the curve approaches them as its branches extend. A student should not insert the asymptote intersection (h,k) as a vertex or as a point of the graph. The centre of a standard hyperbola lies between its two disconnected branches.

Centres and signs: (x+2) does not mean h=+2

The term (x+2)² means (x−(−2))², so its horizontal centre coordinate is h=−2. Similarly, (y−5)² gives k=5. For the equation (x+2)²/16−(y−5)²/9=1, the centre is (−2,5), not (2,−5). Its horizontal vertices are (−6,5) and (2,5), and the asymptote slopes are ±3/4.

Before drawing a curve, write the centre in a separate line. It is faster to catch a reversed sign there than to repair the graph after vertices, asymptotes and range have all inherited the same error. A convenient numerical check is to evaluate the equation at a claimed vertex; only the positive squared term should contribute one at that point.

Why a hyperbola is not an ellipse

An ellipse in standard form has two nonnegative squared terms added to one, such as x²/9+y²/4=1. The resulting curve is closed and bounded. A hyperbola has a difference of squared terms equal to one, so its curve is open with two branches and asymptotes. Changing a minus to a plus creates a different type of conic, not a cosmetic algebraic variation.

The official graph topic includes ellipses and hyperbolas among its standard forms. A learner should distinguish their shapes before trying to use a graphing calculator. The positive-one normalisation makes the difference obvious: sum of two squares for an ellipse, difference of two squares for a hyperbola.

Domain and range come from an inequality

For a horizontal hyperbola, rearrange (x−h)²/a²=1+(y−k)²/b². The right side is at least one, so (x−h)²≥a² and therefore x≤h−a or x≥h+a. There is a gap in permissible x-values between the vertices. But every real y can be paired with some valid x, so the range has no analogous gap.

For a vertical hyperbola, rearrange (y−k)²/b²=1+(x−h)²/a². It follows that y≤k−b or y≥k+b. The x-domain is unrestricted. This is a conceptual check on the sketch: a horizontal hyperbola cannot cross the vertical strip between its vertices, and a vertical hyperbola cannot enter the horizontal strip between its two vertices.

An intercept may not exist

For the horizontal hyperbola (x−1)²/9−(y+2)²/4=1, setting y=0 would give (x−1)²/9−1=1, or (x−1)²=18, so it does have x-intercepts x=1±3√2. But setting x=0 gives 1/9−(y+2)²/4=1, which has no real solution. Therefore there is no y-intercept. One cannot infer the existence of intercepts merely because the branches extend indefinitely.

For a translated hyperbola, use the actual axes y=0 or x=0 if the question asks for intercepts. The vertices are located on the hyperbola’s transverse symmetry axis through its centre, not necessarily on either global coordinate axis. Mixing those reference axes is a common source of false intercept claims.

The difference between an axis of symmetry and an asymptote

The axes of symmetry of these standard shifted hyperbolas are the horizontal line y=k and vertical line x=h through the centre. Their asymptotes are the two diagonal lines y−k=±(b/a)(x−h). The symmetry axes bisect the two branches or their shapes; the asymptotes describe straight-line trends far along the branches. They meet at the same centre but serve different geometric purposes.

A student’s sketch should identify the horizontal and vertical centre lines separately from the diagonal asymptotes. At the horizontal vertices (h±a,k), the tangent directions are vertical for the standard horizontal hyperbola. At the vertical vertices (h,k±b), the tangent directions are horizontal. These facts help confirm the orientation of the drawn branches.

A two-point substitution check using a 3–4–5 identity

For the horizontal form, choose a point with x−h=5a/3 and y−k=4b/3. Then the left side is (25/9)−(16/9)=1, so the point lies on the curve. For the vertical form, choose x−h=4a/3 and y−k=5b/3, giving the same calculation. This produces easy, exact nonvertex points on any of the examples in the clinic section.

The point check is independent of visually reading the diagram. If a candidate point fails the original equation, it is not part of that hyperbola no matter where the plotting software places a rounded marker. Replacing decimal approximations with rational test offsets makes this audit particularly reliable.

One-minute diagnostic for Bukit Timah JC2

Show the learner x²/9−y²/4=1. Ask for branch direction (horizontal), vertices (±3,0), and asymptotes y=±(2/3)x. Then reverse the squares to y²/4−x²/9=1 and ask what changes: the vertices and branch direction, not the asymptote slopes. Finally translate the first equation to (x−1)²/9−(y+2)²/4=1 and ask for centre (1,−2) and right vertex (4,−2). These small changes isolate coefficient, orientation and translation skills.

Thirty-two original hyperbola graphing clinics

The following 32 cases alternate between horizontal and vertical hyperbolas with different centres and positive axis parameters. Each states its exact equation, centre, vertices, asymptotes, domain, range and a verified nonvertex point. These are invented mathematical graphs, not physical property dimensions. Work several before viewing their answers, then check both the vertex and the nonvertex point directly in the equation.

1. Horizontal branches: centre (-2, -3)

Graphing problem. Sketch the exact hyperbola (x+2)²/1 − (y+3)²/4 = 1. Name its centre, branch direction, two vertices, both asymptotes, domain and range. Start by identifying which squared term is positive. Do not assume the asymptotes alone tell you which of the two hyperbola orientations was specified. The exercise uses standard Cartesian coordinates, with no measurements from an actual local site.

Complete worked graph information. Here the centre is (-2,-3), a=1, b=2, and the positive x-square makes the two branches open left and right. The vertices are (-3,-3) and (-1,-3). The asymptotes are y−(-3)=±(2)(x−(-2)). The domain is x≤-3 or x≥-1; the range is all real y. Draw the two symmetry axes through the centre separately from the diagonal asymptotes.

Independent substitution check. A nonvertex point on the curve is (-1/3, -1/3). Its normalised squared terms give 25/9 and 16/9 in the appropriate order, so their difference is exactly one. The positive squared term determines whether the branches are horizontal or vertical. If a question instead gave the same centre and denominators but reversed the squared-term order, the branch orientation and vertex locations would change while the asymptote slopes remain ±2. This is the critical transfer check for this question family.

2. Vertical branches: centre (3, 4)

Graphing problem. Sketch the exact hyperbola (y−4)²/25 − (x−3)²/4 = 1. Name its centre, branch direction, two vertices, both asymptotes, domain and range. Start by identifying which squared term is positive. Do not assume the asymptotes alone tell you which of the two hyperbola orientations was specified. The exercise uses standard Cartesian coordinates, with no measurements from an actual local site.

Complete worked graph information. Here the centre is (3,4), a=2, b=5, and the positive y-square makes the two branches open up and down. The vertices are (3,-1) and (3,9). The asymptotes are y−(4)=±(5/2)(x−(3)). The domain is all real x; the range is y≤-1 or y≥9. Draw the two symmetry axes through the centre separately from the diagonal asymptotes.

Independent substitution check. A nonvertex point on the curve is (17/3, 37/3). Its normalised squared terms give 25/9 and 16/9 in the appropriate order, so their difference is exactly one. Asymptotes meet at the centre but are not the centre lines of symmetry themselves. If a question instead gave the same centre and denominators but reversed the squared-term order, the branch orientation and vertex locations would change while the asymptote slopes remain ±5/2. This is the critical transfer check for this question family.

3. Horizontal branches: centre (-1, 2)

Graphing problem. Sketch the exact hyperbola (x+1)²/9 − (y−2)²/4 = 1. Name its centre, branch direction, two vertices, both asymptotes, domain and range. Start by identifying which squared term is positive. Do not assume the asymptotes alone tell you which of the two hyperbola orientations was specified. The exercise uses standard Cartesian coordinates, with no measurements from an actual local site.

Complete worked graph information. Here the centre is (-1,2), a=3, b=2, and the positive x-square makes the two branches open left and right. The vertices are (-4,2) and (2,2). The asymptotes are y−(2)=±(2/3)(x−(-1)). The domain is x≤-4 or x≥2; the range is all real y. Draw the two symmetry axes through the centre separately from the diagonal asymptotes.

Independent substitution check. A nonvertex point on the curve is (4, 14/3). Its normalised squared terms give 25/9 and 16/9 in the appropriate order, so their difference is exactly one. A horizontal hyperbola has a restricted x-domain and unrestricted y-range. If a question instead gave the same centre and denominators but reversed the squared-term order, the branch orientation and vertex locations would change while the asymptote slopes remain ±2/3. This is the critical transfer check for this question family.

4. Vertical branches: centre (4, 0)

Graphing problem. Sketch the exact hyperbola y²/25 − (x−4)²/16 = 1. Name its centre, branch direction, two vertices, both asymptotes, domain and range. Start by identifying which squared term is positive. Do not assume the asymptotes alone tell you which of the two hyperbola orientations was specified. The exercise uses standard Cartesian coordinates, with no measurements from an actual local site.

Complete worked graph information. Here the centre is (4,0), a=4, b=5, and the positive y-square makes the two branches open up and down. The vertices are (4,-5) and (4,5). The asymptotes are y−(0)=±(5/4)(x−(4)). The domain is all real x; the range is y≤-5 or y≥5. Draw the two symmetry axes through the centre separately from the diagonal asymptotes.

Independent substitution check. A nonvertex point on the curve is (28/3, 25/3). Its normalised squared terms give 25/9 and 16/9 in the appropriate order, so their difference is exactly one. A vertical hyperbola has unrestricted x-domain and a gap in the y-range. If a question instead gave the same centre and denominators but reversed the squared-term order, the branch orientation and vertex locations would change while the asymptote slopes remain ±5/4. This is the critical transfer check for this question family.

5. Horizontal branches: centre (0, -2)

Graphing problem. Sketch the exact hyperbola x²/25 − (y+2)²/4 = 1. Name its centre, branch direction, two vertices, both asymptotes, domain and range. Start by identifying which squared term is positive. Do not assume the asymptotes alone tell you which of the two hyperbola orientations was specified. The exercise uses standard Cartesian coordinates, with no measurements from an actual local site.

Complete worked graph information. Here the centre is (0,-2), a=5, b=2, and the positive x-square makes the two branches open left and right. The vertices are (-5,-2) and (5,-2). The asymptotes are y−(-2)=±(2/5)(x−(0)). The domain is x≤-5 or x≥5; the range is all real y. Draw the two symmetry axes through the centre separately from the diagonal asymptotes.

Independent substitution check. A nonvertex point on the curve is (25/3, 2/3). Its normalised squared terms give 25/9 and 16/9 in the appropriate order, so their difference is exactly one. Changing the centre translates every vertex and asymptote without changing a or b. If a question instead gave the same centre and denominators but reversed the squared-term order, the branch orientation and vertex locations would change while the asymptote slopes remain ±2/5. This is the critical transfer check for this question family.

6. Vertical branches: centre (-4, -4)

Graphing problem. Sketch the exact hyperbola (y+4)²/25 − (x+4)²/1 = 1. Name its centre, branch direction, two vertices, both asymptotes, domain and range. Start by identifying which squared term is positive. Do not assume the asymptotes alone tell you which of the two hyperbola orientations was specified. The exercise uses standard Cartesian coordinates, with no measurements from an actual local site.

Complete worked graph information. Here the centre is (-4,-4), a=1, b=5, and the positive y-square makes the two branches open up and down. The vertices are (-4,-9) and (-4,1). The asymptotes are y−(-4)=±(5)(x−(-4)). The domain is all real x; the range is y≤-9 or y≥1. Draw the two symmetry axes through the centre separately from the diagonal asymptotes.

Independent substitution check. A nonvertex point on the curve is (-8/3, 13/3). Its normalised squared terms give 25/9 and 16/9 in the appropriate order, so their difference is exactly one. The two orientations with matching denominators share asymptotes but not the same branches. If a question instead gave the same centre and denominators but reversed the squared-term order, the branch orientation and vertex locations would change while the asymptote slopes remain ±5. This is the critical transfer check for this question family.

7. Horizontal branches: centre (1, 3)

Graphing problem. Sketch the exact hyperbola (x−1)²/4 − (y−3)²/4 = 1. Name its centre, branch direction, two vertices, both asymptotes, domain and range. Start by identifying which squared term is positive. Do not assume the asymptotes alone tell you which of the two hyperbola orientations was specified. The exercise uses standard Cartesian coordinates, with no measurements from an actual local site.

Complete worked graph information. Here the centre is (1,3), a=2, b=2, and the positive x-square makes the two branches open left and right. The vertices are (-1,3) and (3,3). The asymptotes are y−(3)=±(1)(x−(1)). The domain is x≤-1 or x≥3; the range is all real y. Draw the two symmetry axes through the centre separately from the diagonal asymptotes.

Independent substitution check. A nonvertex point on the curve is (13/3, 17/3). Its normalised squared terms give 25/9 and 16/9 in the appropriate order, so their difference is exactly one. Exact rational test points verify the equation without relying on graphing accuracy. If a question instead gave the same centre and denominators but reversed the squared-term order, the branch orientation and vertex locations would change while the asymptote slopes remain ±1. This is the critical transfer check for this question family.

8. Vertical branches: centre (-3, 1)

Graphing problem. Sketch the exact hyperbola (y−1)²/25 − (x+3)²/9 = 1. Name its centre, branch direction, two vertices, both asymptotes, domain and range. Start by identifying which squared term is positive. Do not assume the asymptotes alone tell you which of the two hyperbola orientations was specified. The exercise uses standard Cartesian coordinates, with no measurements from an actual local site.

Complete worked graph information. Here the centre is (-3,1), a=3, b=5, and the positive y-square makes the two branches open up and down. The vertices are (-3,-4) and (-3,6). The asymptotes are y−(1)=±(5/3)(x−(-3)). The domain is all real x; the range is y≤-4 or y≥6. Draw the two symmetry axes through the centre separately from the diagonal asymptotes.

Independent substitution check. A nonvertex point on the curve is (1, 28/3). Its normalised squared terms give 25/9 and 16/9 in the appropriate order, so their difference is exactly one. The vertices lie a fixed distance a horizontally or b vertically from the centre. If a question instead gave the same centre and denominators but reversed the squared-term order, the branch orientation and vertex locations would change while the asymptote slopes remain ±5/3. This is the critical transfer check for this question family.

9. Horizontal branches: centre (2, -1)

Graphing problem. Sketch the exact hyperbola (x−2)²/16 − (y+1)²/4 = 1. Name its centre, branch direction, two vertices, both asymptotes, domain and range. Start by identifying which squared term is positive. Do not assume the asymptotes alone tell you which of the two hyperbola orientations was specified. The exercise uses standard Cartesian coordinates, with no measurements from an actual local site.

Complete worked graph information. Here the centre is (2,-1), a=4, b=2, and the positive x-square makes the two branches open left and right. The vertices are (-2,-1) and (6,-1). The asymptotes are y−(-1)=±(1/2)(x−(2)). The domain is x≤-2 or x≥6; the range is all real y. Draw the two symmetry axes through the centre separately from the diagonal asymptotes.

Independent substitution check. A nonvertex point on the curve is (26/3, 5/3). Its normalised squared terms give 25/9 and 16/9 in the appropriate order, so their difference is exactly one. The minus sign identifies a hyperbola rather than the plus-form ellipse. If a question instead gave the same centre and denominators but reversed the squared-term order, the branch orientation and vertex locations would change while the asymptote slopes remain ±1/2. This is the critical transfer check for this question family.

10. Vertical branches: centre (-2, -3)

Graphing problem. Sketch the exact hyperbola (y+3)²/25 − (x+2)²/25 = 1. Name its centre, branch direction, two vertices, both asymptotes, domain and range. Start by identifying which squared term is positive. Do not assume the asymptotes alone tell you which of the two hyperbola orientations was specified. The exercise uses standard Cartesian coordinates, with no measurements from an actual local site.

Complete worked graph information. Here the centre is (-2,-3), a=5, b=5, and the positive y-square makes the two branches open up and down. The vertices are (-2,-8) and (-2,2). The asymptotes are y−(-3)=±(1)(x−(-2)). The domain is all real x; the range is y≤-8 or y≥2. Draw the two symmetry axes through the centre separately from the diagonal asymptotes.

Independent substitution check. A nonvertex point on the curve is (14/3, 16/3). Its normalised squared terms give 25/9 and 16/9 in the appropriate order, so their difference is exactly one. An intercept should be computed using x=0 or y=0, not mistaken for a vertex. If a question instead gave the same centre and denominators but reversed the squared-term order, the branch orientation and vertex locations would change while the asymptote slopes remain ±1. This is the critical transfer check for this question family.

11. Horizontal branches: centre (3, 4)

Graphing problem. Sketch the exact hyperbola (x−3)²/1 − (y−4)²/4 = 1. Name its centre, branch direction, two vertices, both asymptotes, domain and range. Start by identifying which squared term is positive. Do not assume the asymptotes alone tell you which of the two hyperbola orientations was specified. The exercise uses standard Cartesian coordinates, with no measurements from an actual local site.

Complete worked graph information. Here the centre is (3,4), a=1, b=2, and the positive x-square makes the two branches open left and right. The vertices are (2,4) and (4,4). The asymptotes are y−(4)=±(2)(x−(3)). The domain is x≤2 or x≥4; the range is all real y. Draw the two symmetry axes through the centre separately from the diagonal asymptotes.

Independent substitution check. A nonvertex point on the curve is (14/3, 20/3). Its normalised squared terms give 25/9 and 16/9 in the appropriate order, so their difference is exactly one. The asymptote slope is b/a under the denominator convention shown in this guide. If a question instead gave the same centre and denominators but reversed the squared-term order, the branch orientation and vertex locations would change while the asymptote slopes remain ±2. This is the critical transfer check for this question family.

12. Vertical branches: centre (-1, 2)

Graphing problem. Sketch the exact hyperbola (y−2)²/25 − (x+1)²/4 = 1. Name its centre, branch direction, two vertices, both asymptotes, domain and range. Start by identifying which squared term is positive. Do not assume the asymptotes alone tell you which of the two hyperbola orientations was specified. The exercise uses standard Cartesian coordinates, with no measurements from an actual local site.

Complete worked graph information. Here the centre is (-1,2), a=2, b=5, and the positive y-square makes the two branches open up and down. The vertices are (-1,-3) and (-1,7). The asymptotes are y−(2)=±(5/2)(x−(-1)). The domain is all real x; the range is y≤-3 or y≥7. Draw the two symmetry axes through the centre separately from the diagonal asymptotes.

Independent substitution check. A nonvertex point on the curve is (5/3, 31/3). Its normalised squared terms give 25/9 and 16/9 in the appropriate order, so their difference is exactly one. Use the positive-one standard form before reading the signs and the branch direction. If a question instead gave the same centre and denominators but reversed the squared-term order, the branch orientation and vertex locations would change while the asymptote slopes remain ±5/2. This is the critical transfer check for this question family.

13. Horizontal branches: centre (4, 0)

Graphing problem. Sketch the exact hyperbola (x−4)²/9 − y²/4 = 1. Name its centre, branch direction, two vertices, both asymptotes, domain and range. Start by identifying which squared term is positive. Do not assume the asymptotes alone tell you which of the two hyperbola orientations was specified. The exercise uses standard Cartesian coordinates, with no measurements from an actual local site.

Complete worked graph information. Here the centre is (4,0), a=3, b=2, and the positive x-square makes the two branches open left and right. The vertices are (1,0) and (7,0). The asymptotes are y−(0)=±(2/3)(x−(4)). The domain is x≤1 or x≥7; the range is all real y. Draw the two symmetry axes through the centre separately from the diagonal asymptotes.

Independent substitution check. A nonvertex point on the curve is (9, 8/3). Its normalised squared terms give 25/9 and 16/9 in the appropriate order, so their difference is exactly one. The positive squared term determines whether the branches are horizontal or vertical. If a question instead gave the same centre and denominators but reversed the squared-term order, the branch orientation and vertex locations would change while the asymptote slopes remain ±2/3. This is the critical transfer check for this question family.

14. Vertical branches: centre (0, -2)

Graphing problem. Sketch the exact hyperbola (y+2)²/25 − x²/16 = 1. Name its centre, branch direction, two vertices, both asymptotes, domain and range. Start by identifying which squared term is positive. Do not assume the asymptotes alone tell you which of the two hyperbola orientations was specified. The exercise uses standard Cartesian coordinates, with no measurements from an actual local site.

Complete worked graph information. Here the centre is (0,-2), a=4, b=5, and the positive y-square makes the two branches open up and down. The vertices are (0,-7) and (0,3). The asymptotes are y−(-2)=±(5/4)(x−(0)). The domain is all real x; the range is y≤-7 or y≥3. Draw the two symmetry axes through the centre separately from the diagonal asymptotes.

Independent substitution check. A nonvertex point on the curve is (16/3, 19/3). Its normalised squared terms give 25/9 and 16/9 in the appropriate order, so their difference is exactly one. Asymptotes meet at the centre but are not the centre lines of symmetry themselves. If a question instead gave the same centre and denominators but reversed the squared-term order, the branch orientation and vertex locations would change while the asymptote slopes remain ±5/4. This is the critical transfer check for this question family.

15. Horizontal branches: centre (-4, -4)

Graphing problem. Sketch the exact hyperbola (x+4)²/25 − (y+4)²/4 = 1. Name its centre, branch direction, two vertices, both asymptotes, domain and range. Start by identifying which squared term is positive. Do not assume the asymptotes alone tell you which of the two hyperbola orientations was specified. The exercise uses standard Cartesian coordinates, with no measurements from an actual local site.

Complete worked graph information. Here the centre is (-4,-4), a=5, b=2, and the positive x-square makes the two branches open left and right. The vertices are (-9,-4) and (1,-4). The asymptotes are y−(-4)=±(2/5)(x−(-4)). The domain is x≤-9 or x≥1; the range is all real y. Draw the two symmetry axes through the centre separately from the diagonal asymptotes.

Independent substitution check. A nonvertex point on the curve is (13/3, -4/3). Its normalised squared terms give 25/9 and 16/9 in the appropriate order, so their difference is exactly one. A horizontal hyperbola has a restricted x-domain and unrestricted y-range. If a question instead gave the same centre and denominators but reversed the squared-term order, the branch orientation and vertex locations would change while the asymptote slopes remain ±2/5. This is the critical transfer check for this question family.

16. Vertical branches: centre (1, 3)

Graphing problem. Sketch the exact hyperbola (y−3)²/25 − (x−1)²/1 = 1. Name its centre, branch direction, two vertices, both asymptotes, domain and range. Start by identifying which squared term is positive. Do not assume the asymptotes alone tell you which of the two hyperbola orientations was specified. The exercise uses standard Cartesian coordinates, with no measurements from an actual local site.

Complete worked graph information. Here the centre is (1,3), a=1, b=5, and the positive y-square makes the two branches open up and down. The vertices are (1,-2) and (1,8). The asymptotes are y−(3)=±(5)(x−(1)). The domain is all real x; the range is y≤-2 or y≥8. Draw the two symmetry axes through the centre separately from the diagonal asymptotes.

Independent substitution check. A nonvertex point on the curve is (7/3, 34/3). Its normalised squared terms give 25/9 and 16/9 in the appropriate order, so their difference is exactly one. A vertical hyperbola has unrestricted x-domain and a gap in the y-range. If a question instead gave the same centre and denominators but reversed the squared-term order, the branch orientation and vertex locations would change while the asymptote slopes remain ±5. This is the critical transfer check for this question family.

17. Horizontal branches: centre (-3, 1)

Graphing problem. Sketch the exact hyperbola (x+3)²/4 − (y−1)²/4 = 1. Name its centre, branch direction, two vertices, both asymptotes, domain and range. Start by identifying which squared term is positive. Do not assume the asymptotes alone tell you which of the two hyperbola orientations was specified. The exercise uses standard Cartesian coordinates, with no measurements from an actual local site.

Complete worked graph information. Here the centre is (-3,1), a=2, b=2, and the positive x-square makes the two branches open left and right. The vertices are (-5,1) and (-1,1). The asymptotes are y−(1)=±(1)(x−(-3)). The domain is x≤-5 or x≥-1; the range is all real y. Draw the two symmetry axes through the centre separately from the diagonal asymptotes.

Independent substitution check. A nonvertex point on the curve is (1/3, 11/3). Its normalised squared terms give 25/9 and 16/9 in the appropriate order, so their difference is exactly one. Changing the centre translates every vertex and asymptote without changing a or b. If a question instead gave the same centre and denominators but reversed the squared-term order, the branch orientation and vertex locations would change while the asymptote slopes remain ±1. This is the critical transfer check for this question family.

18. Vertical branches: centre (2, -1)

Graphing problem. Sketch the exact hyperbola (y+1)²/25 − (x−2)²/9 = 1. Name its centre, branch direction, two vertices, both asymptotes, domain and range. Start by identifying which squared term is positive. Do not assume the asymptotes alone tell you which of the two hyperbola orientations was specified. The exercise uses standard Cartesian coordinates, with no measurements from an actual local site.

Complete worked graph information. Here the centre is (2,-1), a=3, b=5, and the positive y-square makes the two branches open up and down. The vertices are (2,-6) and (2,4). The asymptotes are y−(-1)=±(5/3)(x−(2)). The domain is all real x; the range is y≤-6 or y≥4. Draw the two symmetry axes through the centre separately from the diagonal asymptotes.

Independent substitution check. A nonvertex point on the curve is (6, 22/3). Its normalised squared terms give 25/9 and 16/9 in the appropriate order, so their difference is exactly one. The two orientations with matching denominators share asymptotes but not the same branches. If a question instead gave the same centre and denominators but reversed the squared-term order, the branch orientation and vertex locations would change while the asymptote slopes remain ±5/3. This is the critical transfer check for this question family.

19. Horizontal branches: centre (-2, -3)

Graphing problem. Sketch the exact hyperbola (x+2)²/16 − (y+3)²/4 = 1. Name its centre, branch direction, two vertices, both asymptotes, domain and range. Start by identifying which squared term is positive. Do not assume the asymptotes alone tell you which of the two hyperbola orientations was specified. The exercise uses standard Cartesian coordinates, with no measurements from an actual local site.

Complete worked graph information. Here the centre is (-2,-3), a=4, b=2, and the positive x-square makes the two branches open left and right. The vertices are (-6,-3) and (2,-3). The asymptotes are y−(-3)=±(1/2)(x−(-2)). The domain is x≤-6 or x≥2; the range is all real y. Draw the two symmetry axes through the centre separately from the diagonal asymptotes.

Independent substitution check. A nonvertex point on the curve is (14/3, -1/3). Its normalised squared terms give 25/9 and 16/9 in the appropriate order, so their difference is exactly one. Exact rational test points verify the equation without relying on graphing accuracy. If a question instead gave the same centre and denominators but reversed the squared-term order, the branch orientation and vertex locations would change while the asymptote slopes remain ±1/2. This is the critical transfer check for this question family.

20. Vertical branches: centre (3, 4)

Graphing problem. Sketch the exact hyperbola (y−4)²/25 − (x−3)²/25 = 1. Name its centre, branch direction, two vertices, both asymptotes, domain and range. Start by identifying which squared term is positive. Do not assume the asymptotes alone tell you which of the two hyperbola orientations was specified. The exercise uses standard Cartesian coordinates, with no measurements from an actual local site.

Complete worked graph information. Here the centre is (3,4), a=5, b=5, and the positive y-square makes the two branches open up and down. The vertices are (3,-1) and (3,9). The asymptotes are y−(4)=±(1)(x−(3)). The domain is all real x; the range is y≤-1 or y≥9. Draw the two symmetry axes through the centre separately from the diagonal asymptotes.

Independent substitution check. A nonvertex point on the curve is (29/3, 37/3). Its normalised squared terms give 25/9 and 16/9 in the appropriate order, so their difference is exactly one. The vertices lie a fixed distance a horizontally or b vertically from the centre. If a question instead gave the same centre and denominators but reversed the squared-term order, the branch orientation and vertex locations would change while the asymptote slopes remain ±1. This is the critical transfer check for this question family.

21. Horizontal branches: centre (-1, 2)

Graphing problem. Sketch the exact hyperbola (x+1)²/1 − (y−2)²/4 = 1. Name its centre, branch direction, two vertices, both asymptotes, domain and range. Start by identifying which squared term is positive. Do not assume the asymptotes alone tell you which of the two hyperbola orientations was specified. The exercise uses standard Cartesian coordinates, with no measurements from an actual local site.

Complete worked graph information. Here the centre is (-1,2), a=1, b=2, and the positive x-square makes the two branches open left and right. The vertices are (-2,2) and (0,2). The asymptotes are y−(2)=±(2)(x−(-1)). The domain is x≤-2 or x≥0; the range is all real y. Draw the two symmetry axes through the centre separately from the diagonal asymptotes.

Independent substitution check. A nonvertex point on the curve is (2/3, 14/3). Its normalised squared terms give 25/9 and 16/9 in the appropriate order, so their difference is exactly one. The minus sign identifies a hyperbola rather than the plus-form ellipse. If a question instead gave the same centre and denominators but reversed the squared-term order, the branch orientation and vertex locations would change while the asymptote slopes remain ±2. This is the critical transfer check for this question family.

22. Vertical branches: centre (4, 0)

Graphing problem. Sketch the exact hyperbola y²/25 − (x−4)²/4 = 1. Name its centre, branch direction, two vertices, both asymptotes, domain and range. Start by identifying which squared term is positive. Do not assume the asymptotes alone tell you which of the two hyperbola orientations was specified. The exercise uses standard Cartesian coordinates, with no measurements from an actual local site.

Complete worked graph information. Here the centre is (4,0), a=2, b=5, and the positive y-square makes the two branches open up and down. The vertices are (4,-5) and (4,5). The asymptotes are y−(0)=±(5/2)(x−(4)). The domain is all real x; the range is y≤-5 or y≥5. Draw the two symmetry axes through the centre separately from the diagonal asymptotes.

Independent substitution check. A nonvertex point on the curve is (20/3, 25/3). Its normalised squared terms give 25/9 and 16/9 in the appropriate order, so their difference is exactly one. An intercept should be computed using x=0 or y=0, not mistaken for a vertex. If a question instead gave the same centre and denominators but reversed the squared-term order, the branch orientation and vertex locations would change while the asymptote slopes remain ±5/2. This is the critical transfer check for this question family.

23. Horizontal branches: centre (0, -2)

Graphing problem. Sketch the exact hyperbola x²/9 − (y+2)²/4 = 1. Name its centre, branch direction, two vertices, both asymptotes, domain and range. Start by identifying which squared term is positive. Do not assume the asymptotes alone tell you which of the two hyperbola orientations was specified. The exercise uses standard Cartesian coordinates, with no measurements from an actual local site.

Complete worked graph information. Here the centre is (0,-2), a=3, b=2, and the positive x-square makes the two branches open left and right. The vertices are (-3,-2) and (3,-2). The asymptotes are y−(-2)=±(2/3)(x−(0)). The domain is x≤-3 or x≥3; the range is all real y. Draw the two symmetry axes through the centre separately from the diagonal asymptotes.

Independent substitution check. A nonvertex point on the curve is (5, 2/3). Its normalised squared terms give 25/9 and 16/9 in the appropriate order, so their difference is exactly one. The asymptote slope is b/a under the denominator convention shown in this guide. If a question instead gave the same centre and denominators but reversed the squared-term order, the branch orientation and vertex locations would change while the asymptote slopes remain ±2/3. This is the critical transfer check for this question family.

24. Vertical branches: centre (-4, -4)

Graphing problem. Sketch the exact hyperbola (y+4)²/25 − (x+4)²/16 = 1. Name its centre, branch direction, two vertices, both asymptotes, domain and range. Start by identifying which squared term is positive. Do not assume the asymptotes alone tell you which of the two hyperbola orientations was specified. The exercise uses standard Cartesian coordinates, with no measurements from an actual local site.

Complete worked graph information. Here the centre is (-4,-4), a=4, b=5, and the positive y-square makes the two branches open up and down. The vertices are (-4,-9) and (-4,1). The asymptotes are y−(-4)=±(5/4)(x−(-4)). The domain is all real x; the range is y≤-9 or y≥1. Draw the two symmetry axes through the centre separately from the diagonal asymptotes.

Independent substitution check. A nonvertex point on the curve is (4/3, 13/3). Its normalised squared terms give 25/9 and 16/9 in the appropriate order, so their difference is exactly one. Use the positive-one standard form before reading the signs and the branch direction. If a question instead gave the same centre and denominators but reversed the squared-term order, the branch orientation and vertex locations would change while the asymptote slopes remain ±5/4. This is the critical transfer check for this question family.

25. Horizontal branches: centre (1, 3)

Graphing problem. Sketch the exact hyperbola (x−1)²/25 − (y−3)²/4 = 1. Name its centre, branch direction, two vertices, both asymptotes, domain and range. Start by identifying which squared term is positive. Do not assume the asymptotes alone tell you which of the two hyperbola orientations was specified. The exercise uses standard Cartesian coordinates, with no measurements from an actual local site.

Complete worked graph information. Here the centre is (1,3), a=5, b=2, and the positive x-square makes the two branches open left and right. The vertices are (-4,3) and (6,3). The asymptotes are y−(3)=±(2/5)(x−(1)). The domain is x≤-4 or x≥6; the range is all real y. Draw the two symmetry axes through the centre separately from the diagonal asymptotes.

Independent substitution check. A nonvertex point on the curve is (28/3, 17/3). Its normalised squared terms give 25/9 and 16/9 in the appropriate order, so their difference is exactly one. The positive squared term determines whether the branches are horizontal or vertical. If a question instead gave the same centre and denominators but reversed the squared-term order, the branch orientation and vertex locations would change while the asymptote slopes remain ±2/5. This is the critical transfer check for this question family.

26. Vertical branches: centre (-3, 1)

Graphing problem. Sketch the exact hyperbola (y−1)²/25 − (x+3)²/1 = 1. Name its centre, branch direction, two vertices, both asymptotes, domain and range. Start by identifying which squared term is positive. Do not assume the asymptotes alone tell you which of the two hyperbola orientations was specified. The exercise uses standard Cartesian coordinates, with no measurements from an actual local site.

Complete worked graph information. Here the centre is (-3,1), a=1, b=5, and the positive y-square makes the two branches open up and down. The vertices are (-3,-4) and (-3,6). The asymptotes are y−(1)=±(5)(x−(-3)). The domain is all real x; the range is y≤-4 or y≥6. Draw the two symmetry axes through the centre separately from the diagonal asymptotes.

Independent substitution check. A nonvertex point on the curve is (-5/3, 28/3). Its normalised squared terms give 25/9 and 16/9 in the appropriate order, so their difference is exactly one. Asymptotes meet at the centre but are not the centre lines of symmetry themselves. If a question instead gave the same centre and denominators but reversed the squared-term order, the branch orientation and vertex locations would change while the asymptote slopes remain ±5. This is the critical transfer check for this question family.

27. Horizontal branches: centre (2, -1)

Graphing problem. Sketch the exact hyperbola (x−2)²/4 − (y+1)²/4 = 1. Name its centre, branch direction, two vertices, both asymptotes, domain and range. Start by identifying which squared term is positive. Do not assume the asymptotes alone tell you which of the two hyperbola orientations was specified. The exercise uses standard Cartesian coordinates, with no measurements from an actual local site.

Complete worked graph information. Here the centre is (2,-1), a=2, b=2, and the positive x-square makes the two branches open left and right. The vertices are (0,-1) and (4,-1). The asymptotes are y−(-1)=±(1)(x−(2)). The domain is x≤0 or x≥4; the range is all real y. Draw the two symmetry axes through the centre separately from the diagonal asymptotes.

Independent substitution check. A nonvertex point on the curve is (16/3, 5/3). Its normalised squared terms give 25/9 and 16/9 in the appropriate order, so their difference is exactly one. A horizontal hyperbola has a restricted x-domain and unrestricted y-range. If a question instead gave the same centre and denominators but reversed the squared-term order, the branch orientation and vertex locations would change while the asymptote slopes remain ±1. This is the critical transfer check for this question family.

28. Vertical branches: centre (-2, -3)

Graphing problem. Sketch the exact hyperbola (y+3)²/25 − (x+2)²/9 = 1. Name its centre, branch direction, two vertices, both asymptotes, domain and range. Start by identifying which squared term is positive. Do not assume the asymptotes alone tell you which of the two hyperbola orientations was specified. The exercise uses standard Cartesian coordinates, with no measurements from an actual local site.

Complete worked graph information. Here the centre is (-2,-3), a=3, b=5, and the positive y-square makes the two branches open up and down. The vertices are (-2,-8) and (-2,2). The asymptotes are y−(-3)=±(5/3)(x−(-2)). The domain is all real x; the range is y≤-8 or y≥2. Draw the two symmetry axes through the centre separately from the diagonal asymptotes.

Independent substitution check. A nonvertex point on the curve is (2, 16/3). Its normalised squared terms give 25/9 and 16/9 in the appropriate order, so their difference is exactly one. A vertical hyperbola has unrestricted x-domain and a gap in the y-range. If a question instead gave the same centre and denominators but reversed the squared-term order, the branch orientation and vertex locations would change while the asymptote slopes remain ±5/3. This is the critical transfer check for this question family.

29. Horizontal branches: centre (3, 4)

Graphing problem. Sketch the exact hyperbola (x−3)²/16 − (y−4)²/4 = 1. Name its centre, branch direction, two vertices, both asymptotes, domain and range. Start by identifying which squared term is positive. Do not assume the asymptotes alone tell you which of the two hyperbola orientations was specified. The exercise uses standard Cartesian coordinates, with no measurements from an actual local site.

Complete worked graph information. Here the centre is (3,4), a=4, b=2, and the positive x-square makes the two branches open left and right. The vertices are (-1,4) and (7,4). The asymptotes are y−(4)=±(1/2)(x−(3)). The domain is x≤-1 or x≥7; the range is all real y. Draw the two symmetry axes through the centre separately from the diagonal asymptotes.

Independent substitution check. A nonvertex point on the curve is (29/3, 20/3). Its normalised squared terms give 25/9 and 16/9 in the appropriate order, so their difference is exactly one. Changing the centre translates every vertex and asymptote without changing a or b. If a question instead gave the same centre and denominators but reversed the squared-term order, the branch orientation and vertex locations would change while the asymptote slopes remain ±1/2. This is the critical transfer check for this question family.

30. Vertical branches: centre (-1, 2)

Graphing problem. Sketch the exact hyperbola (y−2)²/25 − (x+1)²/25 = 1. Name its centre, branch direction, two vertices, both asymptotes, domain and range. Start by identifying which squared term is positive. Do not assume the asymptotes alone tell you which of the two hyperbola orientations was specified. The exercise uses standard Cartesian coordinates, with no measurements from an actual local site.

Complete worked graph information. Here the centre is (-1,2), a=5, b=5, and the positive y-square makes the two branches open up and down. The vertices are (-1,-3) and (-1,7). The asymptotes are y−(2)=±(1)(x−(-1)). The domain is all real x; the range is y≤-3 or y≥7. Draw the two symmetry axes through the centre separately from the diagonal asymptotes.

Independent substitution check. A nonvertex point on the curve is (17/3, 31/3). Its normalised squared terms give 25/9 and 16/9 in the appropriate order, so their difference is exactly one. The two orientations with matching denominators share asymptotes but not the same branches. If a question instead gave the same centre and denominators but reversed the squared-term order, the branch orientation and vertex locations would change while the asymptote slopes remain ±1. This is the critical transfer check for this question family.

31. Horizontal branches: centre (4, 0)

Graphing problem. Sketch the exact hyperbola (x−4)²/1 − y²/4 = 1. Name its centre, branch direction, two vertices, both asymptotes, domain and range. Start by identifying which squared term is positive. Do not assume the asymptotes alone tell you which of the two hyperbola orientations was specified. The exercise uses standard Cartesian coordinates, with no measurements from an actual local site.

Complete worked graph information. Here the centre is (4,0), a=1, b=2, and the positive x-square makes the two branches open left and right. The vertices are (3,0) and (5,0). The asymptotes are y−(0)=±(2)(x−(4)). The domain is x≤3 or x≥5; the range is all real y. Draw the two symmetry axes through the centre separately from the diagonal asymptotes.

Independent substitution check. A nonvertex point on the curve is (17/3, 8/3). Its normalised squared terms give 25/9 and 16/9 in the appropriate order, so their difference is exactly one. Exact rational test points verify the equation without relying on graphing accuracy. If a question instead gave the same centre and denominators but reversed the squared-term order, the branch orientation and vertex locations would change while the asymptote slopes remain ±2. This is the critical transfer check for this question family.

32. Vertical branches: centre (0, -2)

Graphing problem. Sketch the exact hyperbola (y+2)²/25 − x²/4 = 1. Name its centre, branch direction, two vertices, both asymptotes, domain and range. Start by identifying which squared term is positive. Do not assume the asymptotes alone tell you which of the two hyperbola orientations was specified. The exercise uses standard Cartesian coordinates, with no measurements from an actual local site.

Complete worked graph information. Here the centre is (0,-2), a=2, b=5, and the positive y-square makes the two branches open up and down. The vertices are (0,-7) and (0,3). The asymptotes are y−(-2)=±(5/2)(x−(0)). The domain is all real x; the range is y≤-7 or y≥3. Draw the two symmetry axes through the centre separately from the diagonal asymptotes.

Independent substitution check. A nonvertex point on the curve is (8/3, 19/3). Its normalised squared terms give 25/9 and 16/9 in the appropriate order, so their difference is exactly one. The vertices lie a fixed distance a horizontally or b vertically from the centre. If a question instead gave the same centre and denominators but reversed the squared-term order, the branch orientation and vertex locations would change while the asymptote slopes remain ±5/2. This is the critical transfer check for this question family.

Quick answer checks

x²/9−y²/4=1: horizontal branches, vertices (−3,0), (3,0), asymptotes y=±2x/3.
y²/4−x²/9=1: vertical branches, vertices (0,−2), (0,2), asymptotes y=±2x/3.
(x−1)²/9−(y+2)²/4=1: centre (1,−2), vertices (−2,−2) and (4,−2).
The corresponding shifted vertical form has vertices (1,−4) and (1,0).
For x²/9−y²/4=1, the x-domain is x≤−3 or x≥3.
For y²/4−x²/9=1, the y-range is y≤−2 or y≥2.
At the centre (h,k), the hyperbola expression equals 0, not 1.
A nonvertex point with offsets 5a/3 and 4b/3 from the centre satisfies the horizontal form.
Hyperbola asymptotes are diagonal under these standard nondegenerate equations.
A plus sign between both squared terms defines an ellipse, not a hyperbola, in these standard forms.

A five-day study route

Day one: compare the standard ellipse and two hyperbola equations and identify which squared term is positive. Day two: extract centres and vertices from translated forms, checking the bracket signs. Day three: draw asymptote lines and symmetry axes separately. Day four: derive domain or range restrictions from inequalities and decide whether requested coordinate-axis intercepts exist. Day five: complete mixed cases without headings announcing the orientation, then verify a nonvertex point using the original equation.

In a three-student tutorial, one learner can identify the centre and orientation, another can calculate vertices and asymptotes, and a third can check domain, range and an exact point. Rotating roles makes a wrong reading of the positive squared term visible before the whole sketch is drawn. A strong answer is a labelled, mathematically consistent graph, not just a memorised pair of diagonal lines.

Frequently asked questions

Which direction does a hyperbola open?

When the x-square is positive in the standard difference-of-squares equation, the branches open left and right. When the y-square is positive, they open up and down.

Are hyperbola asymptotes the same as its symmetry axes?

No. The symmetry axes are horizontal and vertical lines through the centre; the asymptotes are diagonal lines through the same centre.

Can two different hyperbolas share their asymptotes?

Yes. The horizontal and vertical standard forms with the same centre and denominators have the same asymptotes but different vertex locations and branches.

Is the centre on a hyperbola?

Not for these standard nondegenerate forms with right side one. Substituting the centre gives zero on the left.

What do a and b represent?

Under this notation, a is the horizontal scale appearing below the x-square and b is the vertical scale appearing below the y-square. The positive term determines which scale reaches the vertices.

Does a hyperbola always cross both coordinate axes?

No. Translating it can remove its intersections with x=0 or y=0. Substitute each axis equation to test for real solutions.

Can a graphing calculator show the exact asymptotes?

It can help visualise them, but the exact asymptote equations follow from the standard algebraic form and should be stated when requested.

Is this part of H2 Mathematics 9758?

Yes. The 2027 syllabus includes hyperbola graphs, their main features and asymptotes in Functions and Graphs.

JC2 H2 Mathematics Tutor Bukit Timah | Hyperbola Graphs: Which Axis Holds the Branches?