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Mathematics in Bukit Timah | Hyperbolas, Distance Differences and Asymptotes

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by

eduKate Asia
Original conceptual illustration showing a linear-programming feasible polygon and multiple points along the same optimal boundary.

Did you know that two fixed points can generate not only an ellipse but also a hyperbola, simply by changing the distance rule? An ellipse keeps the sum of distances constant. A hyperbola keeps their absolute difference constant. Fix imaginary foci F₁=(−5,0) and F₂=(5,0). The set of points whose focal distances differ by six units forms two branches of the hyperbola x²/9−y²/16=1.

This Mathematics in Bukit Timah guide explains focal distance differences, vertices, transverse and conjugate axes, standard and translated equations, asymptotes, eccentricity, tangents, point classification and the limits of geometric modelling. All coordinates are fictional educational constructions, not actual Bukit Timah locations, signals, paths, school data or engineering measurements. This is higher analytic geometry enrichment built from distance, surds, quadratics and graphing.

Quick answer: A horizontal hyperbola x²/a²−y²/b²=1, with positive a,b, has centre at the origin, vertices (±a,0), foci (±c,0) where c²=a²+b², and diagonal asymptotes y=±(b/a)x. For our example a=3,b=4,c=5, the asymptotes are y=±4x/3, and the constant absolute difference of distances to the foci is 2a=6.

Article contents

Original 2026 mathematical plotting illustration used as conceptual support for imaginary hyperbola branches and asymptote geometry; not a literal surveyed graph.

The idea begins with a distance comparison

Given two fixed distinct points, take a moving point P.

Measure its distance to the first focus and its distance to the second focus. Find the absolute difference, which is never negative.

If that difference equals a permitted positive constant, the possible points form two disconnected hyperbola branches. The constant must be less than the focal separation for a nondegenerate ordinary hyperbola.

Why Bukit Timah is context rather than survey evidence

A student near Beauty World, Sixth Avenue or King Albert Park can investigate coordinate curves on a graph.

But no coordinate in this guide locates a property, MRT entrance, route or classroom. We use an ideal Euclidean plane and exact artificial lengths.

A hyperbola model can be applied in other fields, but factual claims about physical objects or travel times require independent observations and appropriate units.

Contrast the ellipse and hyperbola definitions

For an ellipse with two foci, PF₁+PF₂ is fixed.

For a hyperbola, |PF₁−PF₂| is fixed.

The change from sum to absolute difference is decisive. Using the ellipse focal relation with a hyperbola equation will lead to incorrect centre distances and asymptotes.

Worked example 1: identify the foci

Our foci are F₁=(−5,0) and F₂=(5,0).

Their separation is ten units, and their midpoint is (0,0).

The focal half-separation is c=5. The prescribed absolute distance difference is six, so 2a=6 and a=3. Since a<c, a genuine two-branched hyperbola is possible.

The centre is not a point on the curve

The centre lies midway between the foci and between the two branches.

For our centred hyperbola, C=(0,0). Substituting x=y=0 into x²/9−y²/16=1 gives zero, not one.

Therefore the centre does not lie on the curve. It is a symmetry point, not an intersection point.

The vertices are nearest to the centre

The two horizontal vertices are (−a,0) and (a,0).

For a=3, they are (−3,0) and (3,0). These are the points on the branches closest to the centre along the transverse axis.

Do not confuse vertices at ±3 with foci at ±5. Both pairs are on the horizontal axis but perform different geometric roles.

Worked example 2: test the right vertex

Take P=(3,0).

Distance to F₁=(−5,0) is eight. Distance to F₂=(5,0) is two.

Their absolute difference is |8−2|=6, exactly 2a. Therefore the vertex satisfies the defining distance rule.

The transverse axis joins the vertices

For a horizontal hyperbola, the transverse axis is along the x-direction.

The two vertices are separated by 2a units. The branches open outward from those vertices, toward x≤−a and x≥a.

The term transverse axis does not imply the hyperbola contains every point of the segment between its vertices. In fact, that central gap contains no curve points.

The conjugate parameter b does not create a minor-axis boundary

The standard form uses b² in the denominator of the subtracted y² term.

The value b controls how steep the asymptotes are relative to a. It helps define a guiding rectangle of half-height b and half-width a.

Unlike the ellipse’s semi-minor axis, b is not the coordinate of a top or bottom point on the hyperbola along x=0; the curve does not cross that axis.

The standard horizontal equation

For a,b>0, write

x²/a²−y²/b²=1.

The positive term indicates the transverse axis. Because x²/a²=1+y²/b²≥1, any real point on the curve satisfies |x|≥a.

This algebra explains the two branches and the empty central strip −a<x<a.

Worked example 3: find the allowed x-values

For x²/9−y²/16=1, rearrange to x²/9=1+y²/16.

The right side is at least one, so x²≥9, giving x≤−3 or x≥3.

No curve point can have x=0, x=1 or x=−2. A graph that crosses the central vertical axis is not this hyperbola.

The focal relation uses a plus sign

A horizontal ellipse satisfies c²=a²−b². A hyperbola instead satisfies

c²=a²+b².

For a=3,b=4, c=√(9+16)=5.

Thus the foci lie farther from the centre than the vertices, which is consistent with the focal points ±5 and vertices ±3.

Worked example 4: recover b from foci and vertices

Suppose the half focal separation is c=5 and transverse semi-axis length is a=3.

Then b²=c²−a²=25−9=16, giving b=4.

The equation is x²/9−y²/16=1. Choosing minus in the focal relation would incorrectly give a negative b² for this example.

A point on a branch has two possible y-values

Fix a permitted positive x, such as x=5.

Substitute into x²/9−y²/16=1:

25/9−y²/16=1, so y²/16=16/9.

Thus y=±16/3. Both points lie on the right branch, one above and one below the horizontal axis.

Worked example 5: check the upper point

Let P=(5,16/3).

Then x²/9=25/9. Its y²/16 is (256/9)/16=16/9.

The difference is 25/9−16/9=9/9=1, so the point lies on the hyperbola exactly.

Verify that point using both focal distances

For P=(5,16/3), distance to F₂=(5,0) is 16/3.

Distance to F₁=(−5,0) is √[10²+(16/3)²]=√[(900+256)/9]=√(1156/9)=34/3.

Their absolute difference is 34/3−16/3=18/3=6, matching the defining constant. This is independent verification of the coordinate equation.

Why the difference is an absolute value

On the right branch, distance to the left focus tends to exceed distance to the right focus.

On the left branch, that ordering reverses. Taking absolute value covers both branches in one geometric condition.

If a signed difference were fixed instead, the orientation could select only one branch. The conventional hyperbola definition usually uses the absolute difference.

Two branches reflect across the axes

The equation contains x² and y², so changing x to −x or y to −y leaves it unchanged.

Therefore the graph has symmetry about both coordinate axes and the origin.

The point (5,16/3) has companion points (5,−16/3), (−5,16/3), and (−5,−16/3), all satisfying the same equation.

The asymptotes describe far-away direction

An asymptote is a line that the hyperbola approaches at large distances along its branches in the appropriate sense.

For x²/a²−y²/b²=1, the oblique asymptotes are

y=(b/a)x and y=−(b/a)x.

Their slopes are determined by the ratio b/a, not by the focal half-separation c alone.

Worked example 6: find the asymptotes

For x²/9−y²/16=1, a=3 and b=4.

Therefore the two asymptotes are y=4x/3 and y=−4x/3.

Both pass through the origin, the centre. They are guide lines that the branches approach; they are not actual hyperbola branches.

The asymptotes come from large x

Rearrange the right-branch equation to

y=±(b/a)√(x²−a²) for x≥a.

For large positive x, the ratio √(x²−a²)/x approaches one.

Thus y/x approaches ±b/a as x grows. This proves the slope of each asymptotic direction rather than just guessing from a sketch.

Worked example 7: a point farther out

Let x=15 in our hyperbola.

Then y²/16=225/9−1=25−1=24, so y=±4√24=±8√6≈±19.596.

The upper asymptote predicts y=(4/3)(15)=20. The curve point lies slightly below it. The relative difference becomes smaller farther out.

A branch never crosses its own corresponding asymptote

For the upper right branch, y=(b/a)√(x²−a²), while the positive-slope asymptote is y=(b/a)x for x≥a.

Since √(x²−a²)<x when x>a>0, the branch lies strictly below the corresponding asymptote in that quadrant.

The distinction remains even though their vertical separation tends toward zero as x increases.

Worked example 8: compare x equals thirty

For a=3,b=4,x=30, the upper branch has y=(4/3)√(900−9)=(4/3)√891.

The asymptote has y=40.

Because √891<30, the exact hyperbola y-value is less than forty. A calculator is not needed to establish the sign of the gap.

The rectangle behind the asymptotes

Draw the rectangle with corners (±a,±b), which for the example are (±3,±4).

Its diagonals lie on the asymptote lines of slopes ±b/a.

This guiding rectangle helps draw the hyperbola. Its four corners are not generally points on the curve, so a student must not confuse the rectangle with the hyperbola boundary.

Worked example 9: test a guiding-rectangle corner

Take (3,4), one corner of the guide rectangle.

Substitute into x²/9−y²/16: 9/9−16/16=1−1=0, not one.

Therefore (3,4) does not lie on the hyperbola. It helps construct an asymptote direction instead.

Eccentricity is greater than one

For a horizontal hyperbola, define eccentricity e=c/a, where c²=a²+b².

Because b>0, c>a and therefore e>1.

This distinguishes a nondegenerate hyperbola from an ellipse, whose eccentricity lies below one, and a parabola, whose eccentricity is one in the standard conic ratio definition.

Worked example 10: calculate eccentricity

For a=3,b=4,c=5, the eccentricity is e=5/3≈1.667.

It has no physical length unit because both c and a are lengths.

Reporting eccentricity five because the focus is at x=5 confuses a location parameter with a dimensionless ratio.

The focal distance constant is 2a, not 2c

The difference between focal distances for any curve point equals 2a=6.

The separation between the two foci is 2c=10.

Those are not equal. The difference constant must be smaller than the focal separation for a nondegenerate hyperbola.

Worked example 11: compare the two lengths

Our focal separation is ten, while the required absolute distance difference is six.

At right vertex (3,0), the focal distances are eight and two, differing by six.

If someone used ten as the difference constant, the ordinary two-branch hyperbola would collapse to a degenerate ray structure rather than the stated curve.

Translation shifts the centre and asymptotes

For a horizontal hyperbola centred at (h,k),

(x−h)²/a²−(y−k)²/b²=1.

Its vertices are (h±a,k), foci (h±c,k), and asymptotes y−k=±(b/a)(x−h).

Translation changes absolute locations but not a,b,c or eccentricity.

Worked example 12: shift right and down

Consider (x−2)²/9−(y+1)²/16=1.

Centre is (2,−1), vertices are (−1,−1) and (5,−1), and foci are (−3,−1) and (7,−1).

The asymptotes pass through (2,−1), with equations y+1=±(4/3)(x−2).

Vertical hyperbolas reverse which square is positive

For a hyperbola whose branches open up and down,

y²/a²−x²/b²=1.

Its vertices are (0,±a), foci (0,±c), and asymptotes y=±(a/b)x.

The positive square term identifies the transverse-axis direction. Do not force the horizontal formulas onto a vertical hyperbola without swapping the roles of x and y.

Worked example 13: a vertical hyperbola

Take y²/16−x²/9=1.

Then a=4,b=3,c=√(16+9)=5.

Its vertices are (0,±4), its foci are (0,±5), and its asymptotes are y=±4x/3.

A vertical hyperbola cannot cross the horizontal axis

From y²/16−x²/9=1, we have y²/16=1+x²/9≥1.

Therefore |y|≥4 for every curve point.

There are no points with y=0. The upper branch begins at (0,4) and the lower at (0,−4).

Worked example 14: find points at x equals three

Set x=3 in y²/16−x²/9=1.

Then y²/16−1=1, so y²=32.

Thus y=±4√2. Both points lie on the vertical branches, one above and one below the centre.

Asymptotes can be the same for conjugate hyperbolas

The horizontal hyperbola x²/9−y²/16=1 has asymptotes y=±4x/3.

The vertical hyperbola y²/16−x²/9=1 has those same asymptotes.

These are conjugate hyperbolas: they share the centre and asymptote directions, but their branches open along different axes. Matching asymptotes does not imply identical curves.

Tangent lines can be found implicitly

For x²/a²−y²/b²=1, differentiate with respect to x:

2x/a²−(2y/b²)(dy/dx)=0.

Where y≠0, this gives dy/dx=b²x/(a²y).

At a vertex where y=0, the tangent may instead be vertical, and the derivative expression cannot be used as a finite slope.

Worked example 15: tangent at five, sixteen-thirds

Use x²/9−y²/16=1 and P=(5,16/3).

Slope dy/dx=(16×5)/(9×16/3)=80/48=5/3.

A tangent through P is y−16/3=(5/3)(x−5). Simplifying gives 5x−3y=9.

The standard tangent equation

For a point (x₀,y₀) on the horizontal hyperbola,

xx₀/a²−yy₀/b²=1

is its tangent line.

For x₀=5,y₀=16/3,a²=9,b²=16, the equation becomes 5x/9−y/3=1, equivalent to 5x−3y=9.

This shortcut also handles the vertices without division by y.

Worked example 16: tangent at the right vertex

At (3,0), the tangent formula gives 3x/9−0=1, so x=3.

This is a vertical line. The graph has a leftmost point on its right branch at x=3, and the tangent is vertical there.

A vertical tangent is not the same as a missing or nonexistent tangent.

Verify a tangent through repeated roots

Substitute y=(5x−9)/3 into x²/9−y²/16=1.

The algebra reduces to a quadratic whose intersection at x=5 is a repeated root, reflecting tangency.

An independent check is that the line has gradient 5/3 and passes through the verified on-curve point (5,16/3).

Worked example 17: a line cutting the hyperbola

Set y=0 in x²/9−y²/16=1.

Then x²/9=1, giving x=±3.

Thus the horizontal line y=0 intersects both branches at the two vertices (−3,0) and (3,0). It is a secant, not an asymptote.

A hyperbola is unbounded

Unlike an ellipse, the two branches extend indefinitely.

For large |x| on the horizontal hyperbola, corresponding |y| values also become large, approaching the asymptotes.

This does not mean a real road or curved structure has infinite length. It is a property of the ideal mathematical locus with no physical boundaries.

Worked example 18: a bounded viewing window

Suppose we plot only x-values from −10 to ten and y-values from −10 to ten.

The displayed portions of the hyperbola are finite because the viewing window is finite.

The curve itself remains unbounded outside the window. A graph-paper crop is not an additional equation defining the whole hyperbola.

Absolute focal difference has a geometric bound

For any three points P,F₁,F₂, the reverse triangle inequality gives

|PF₁−PF₂|≤F₁F₂.

Equality occurs in certain collinear configurations. For a nondegenerate hyperbola the fixed difference is strictly between zero and the focal separation.

This inequality helps check whether a proposed pair of foci and distance-difference constant can define the intended curve.

Worked example 19: reject an impossible difference

Let F₁F₂=10 units.

A request for a point P with |PF₁−PF₂|=12 cannot be met in Euclidean space because the absolute difference cannot exceed ten.

Therefore no such point exists. The problem fails before any coordinate algebra is attempted.

Why the branches begin beyond the vertices

For our horizontal equation, every point obeys |x|≥3.

At y=0, the boundary values are x=±3. At nonzero y, x²/9=1+y²/16>1, so |x|>3.

Thus the two vertices are closest to the centre along the x-direction, and the branches move outward as their vertical magnitude grows.

Worked example 20: solve x as a function of y

Rearrange x²/9−y²/16=1:

x²/9=1+y²/16.

Then x=±3√(1+y²/16). The plus sign describes the right branch and the minus sign the left branch.

For y=0 we get x=±3. For any nonzero y, the absolute x-coordinate exceeds three.

Why a hyperbola is not a single-valued y function globally

For many permitted x-values there are two y-values, one positive and one negative.

For example x=5 gives y=±16/3. Therefore the complete hyperbola fails the vertical-line test for being one function y=f(x).

An individual upper or lower branch segment can be represented as a function on an appropriate restricted domain.

The inverse function idea has its own hyperbola

A familiar hyperbola is xy=k, equivalently y=k/x for nonzero x.

It has coordinate-axis asymptotes x=0 and y=0, and it is a rectangular hyperbola in a rotated coordinate orientation.

This is related to the conic family but not already in the same axis-aligned x²/a²−y²/b² standard form.

Worked example 21: inverse variation

Let y=12/x for x>0.

At x=3, y=4; at x=6, y=2; at x=12, y=1. The product xy remains twelve.

The curve approaches the axes as x becomes very large or approaches zero from the positive side, but it never reaches either axis in the permitted domain.

Rectangular hyperbolas have perpendicular asymptotes

For x²−y²=1, the asymptotes are y=x and y=−x.

Their slopes +1 and −1 multiply to −1, so the asymptotes are perpendicular.

This is one example of a rectangular hyperbola. The condition a=b in the standard horizontal form gives perpendicular asymptotes.

Worked example 22: a rectangular hyperbola point

Take x=√2 and y=1.

Then x²−y²=2−1=1, so the point lies on the hyperbola.

Its asymptotes are y=±x, not the coordinate axes. An appropriate rotation connects rectangular hyperbolas to inverse-variation forms.

Hyperbolic sine and cosine provide a parametrisation

The hyperbolic functions satisfy cosh²t−sinh²t=1.

Therefore setting x=a cosh t and y=b sinh t gives x²/a²−y²/b²=1 for the right branch.

This is an advanced algebraic parametrisation. The functions cosh and sinh are not ordinary cos and sin and should not be interchanged.

Worked example 23: a parametrised vertex

For t=0, cosh0=1 and sinh0=0.

Thus x=a and y=0, giving the right vertex (a,0). The parameter then moves along the right branch as t varies.

An analogous choice x=−a cosh t can describe the left branch under a consistent y parametrisation.

A hyperbola has no finite enclosed area by itself

An ellipse is a closed curve and bounds a finite filled region.

A standard hyperbola consists of two open unbounded branches. It does not alone enclose one finite planar area in the same way.

A finite area question needs additional boundaries such as lines or an interval of integration. Students should not apply ellipse area πab to a hyperbola.

Worked example 24: area requires a region

Suppose a problem asks for the area between the right branch, x-axis and the vertical lines x=3 and x=5.

Those extra boundaries specify a finite region, so an integral of an appropriate y(x) expression could be used.

Without the vertical limits, “area enclosed by the hyperbola” is incomplete for the ordinary open curve.

The conic eccentricity classification

For ellipse e<1, parabola e=1, and hyperbola e>1 under the usual nondegenerate focus-directrix ratio convention.

Our hyperbola’s e=5/3, placing it in the hyperbolic case.

These thresholds distinguish geometric families, but a specific plotted quadratic equation must still be analysed correctly rather than classified from one visual impression.

Worked example 25: distinguish three equations

Ellipse: x²/25+y²/16=1, a closed curve.

Hyperbola: x²/9−y²/16=1, two unbounded branches.

Parabola: x²=8y, one connected unbounded curve with a focus and directrix.

The signs and arrangement of squared terms help distinguish their standard coordinate forms.

Two-dimensional intersections remain algebraic questions

A line can have zero, one, two or in certain special coincident interpretations more shared points with a conic, depending on the equations.

For an ordinary hyperbola and a generic line, substitution produces a quadratic.

The number of real roots controls the ordinary distinct intersections. A diagram can assist but should not replace solving or checking the equation.

The graph’s apparent asymptote may be misleading on a screen

If a plot has different horizontal and vertical visual scales, the asymptote angle can appear steeper or shallower than its geometric coordinate slope implies.

For y=4x/3, the mathematical rise/run is exactly 4/3 in equal coordinate units.

Graphing software should display scale labels, and physical angles should be interpreted only if axis scales are consistent.

The difference-of-distances model can describe timing differences

Under a simplified model with two fixed sources and identical constant signal propagation speed, a constant difference of arrival times corresponds to a constant difference in distances.

That gives hyperbolic loci under appropriate two-dimensional assumptions.

But using such a model for a real system requires source coordinates, propagation physics, synchronized timing and measurement uncertainty. No actual Bukit Timah signal data are available in this lesson.

Worked example 26: a purely hypothetical timing rule

Suppose two fictional sources emit synchronized signals at constant speed v, and the arrival-time difference magnitude is Δt.

Under the ideal assumptions, the distance difference magnitude is vΔt.

If vΔt=6 coordinate-length units and the sources are at (±5,0), the possible receiving positions satisfy the same focal-difference rule as our hyperbola. This is a mathematical mapping, not a measured location.

Common error 1: use the ellipse focal formula

A pupil writes c²=a²−b² for x²/9−y²/16=1.

That gives c²=9−16, which is negative, contradicting the foci in this real horizontal hyperbola.

The correct relation is c²=a²+b²=25, giving c=5.

Common error 2: swap a and b because sixteen is larger

For x²/9−y²/16=1, the positive x² term identifies the transverse direction.

Thus a²=9 and b²=16, even though b>a.

An ellipse places the larger denominator under its major direction; that shortcut cannot be copied into hyperbola problems.

Common error 3: mistake asymptotes for the actual curve

The point (3,4) lies on the line y=4x/3.

But it gives x²/9−y²/16=1−1=0, not one. It is not on the hyperbola.

The asymptote describes limiting direction and is a separate geometric line.

Common error 4: call the focus a vertex

Our vertices are (±3,0) and foci are (±5,0).

Confusing these pairs changes the prescribed focal-distance difference and eccentricity.

Write the ordered quantities a,b,c before plotting the coordinates.

Common error 5: assume the curve crosses the centre

The centre (0,0) gives left side zero, so it is not on x²/9−y²/16=1.

There is an empty central region between the two branches.

The centre is a symmetry reference, not an automatic boundary point.

Common error 6: use the tangent gradient formula at a vertex

The formula dy/dx=b²x/(a²y) divides by y, so it is undefined at the horizontal vertices y=0.

The vertical tangent still exists. At (3,0), its equation is x=3.

Use the tangent identity or geometric reasoning when a slope is infinite or undefined as a finite number.

Common error 7: report an area πab for the hyperbola

The ellipse formula A=πab applies to a closed ellipse region.

A hyperbola has unbounded branches and does not enclose the same kind of finite area on its own.

A hyperbola area problem must provide additional boundaries.

Primary foundations: differences of distances

Younger learners can compare distances from a point to two fixed markers and note that the difference changes with location.

They do not need conic notation to practise reliable subtraction and absolute value.

The later hyperbola definition becomes meaningful only after coordinate distance, Pythagoras and algebraic inequalities are secure.

Secondary progression: graphs and signed terms

Students can revisit coordinate equations, simultaneous equations, surds and graph symmetry.

The key step is recognising that a subtracted square in a standard hyperbola equation creates a different geometry from an ellipse’s sum of squares.

Advanced tangents and asymptotic limits should be taught when the learner’s algebra and calculus background permits them.

Advanced course scope

Hyperbolas, their focus-directrix definitions and advanced parametric forms are conic-section enrichment.

Use the SEAB 2027 G3 Mathematics and Additional Mathematics directory and 2027 A-Level syllabus directory to match actual requirements.

The guide does not assert that every hyperbolic-function and tangent result is mandatory in SEC G3 Mathematics.

A twenty-minute parent diagnostic

Minutes 1–5: Locate foci (−5,0),(5,0), calculate their separation ten and identify the distance-difference constant six.

Minutes 6–10: Use a=3,c=5 and c²=a²+b² to find b=4. Write x²/9−y²/16=1.

Minutes 11–15: Verify (3,0) and (5,16/3) by equation and focal distances.

Minutes 16–20: Find asymptotes y=±4x/3 and explain why (3,4) lies on an asymptote but not on the actual hyperbola.

A seven-day learning route

Day one compares ellipse sums and hyperbola differences. Day two uses focal coordinates and vertices.

Day three derives and checks a,b,c. Day four plots branches and asymptotes.

Day five studies shifted and vertical hyperbolas. Day six finds tangents and distinguishes vertical-gradient cases. Day seven classifies unfamiliar conics by equation, locus and eccentricity.

Practice questions with checked answers

A. For x²/9−y²/16=1, find a,b,c. Answer: 3,4,5.

B. Foci? Answer: (±5,0).

C. Vertices? Answer: (±3,0).

D. Focal distance difference? Answer: 6.

E. Asymptotes? Answer: y=±4x/3.

F. Eccentricity? Answer: 5/3.

G. Tangent at (5,16/3)? Answer: 5x−3y=9.

H. Centre of (x−2)²/9−(y+1)²/16=1? Answer: (2,−1).

Frequently asked questions

What defines a hyperbola?

The nondegenerate hyperbola is the locus where the absolute difference of distances to two fixed foci equals a positive constant smaller than the focal separation.

How does it differ from an ellipse?

The ellipse fixes a distance sum; the hyperbola fixes an absolute distance difference.

Why is c²=a²+b²?

The hyperbola’s focal half-separation c exceeds its transverse semi-axis a, and its geometric derivation yields that plus-sign relation.

Are asymptotes part of the hyperbola?

No. They are separate lines approached by the branches at large distances.

Why are there two branches?

The equation requires |x|≥a in horizontal standard form, leaving a central gap.

Is the centre on the curve?

No. For a standard nondegenerate hyperbola the centre does not satisfy the equation.

What is its eccentricity?

For the horizontal form, e=c/a>1.

Can a hyperbola have a vertical tangent?

Yes, including at the horizontal form’s vertices.

Do these curves represent real Bukit Timah locations?

No. They are fictional coordinate constructions with no mapped travel information.

How can parents recognise understanding?

The learner can distinguish the two focal conditions, calculate c using the correct sign, check a curve point and derive the asymptote slopes without copying an ellipse formula.

Sources and connected Mathematics in Bukit Timah

The SEAB 2027 G3 Mathematics and Additional Mathematics syllabus directory provides Singapore course context; this article’s extended conic mathematics is not asserted to be universally examinable at Secondary level.

Continue with Ellipses and Two Foci, Parabola Focus and Directrix, Coordinate Circle Equations and Tangency, Quadratic Graphs and Vertices and Surds and Exact Distances.

The immutable Clementi Secondary 1 Mathematics small-group reference anchors rigorous, accessible explanation, diagnosed errors and independent verification.

The lasting habit is to state the focal rule, identify the positive squared term, use c²=a²+b², verify an actual curve point and treat asymptotes as limiting lines rather than the curve itself.

Companion Mathematics in Bukit Timah guides in the new four-article batch: Ellipses, Two Foci, Distance Sums and Major Axes · Fractals, Self-Similarity and Fractional Dimension · Lagrange Interpolation, Polynomials and Data Fitting. Together these new lessons extend coordinate loci, conic geometry, recursion and polynomial reconstruction with exact fictional worked examples, careful assumptions and independent checks.