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Mathematics in Bukit Timah | Ellipses, Two Foci, Distance Sums and Major Axes

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by

eduKate Asia
Mathematics textbooks stand behind an open notebook containing written working and a graph, with a calculator and pens alongside.

Did you know that an ellipse can be drawn using two fixed points and one constant distance sum? Put two pins at F₁=(−3,0) and F₂=(3,0) on imaginary graph paper. Imagine a string that makes the sum of the distances from a moving pencil to the pins equal ten units. The resulting curve has equation x²/25+y²/16=1. This connection between a geometric rule and an algebraic equation is one of the most elegant ideas in coordinate geometry.

This Mathematics in Bukit Timah guide explains the two-focus definition of an ellipse, centre, major and minor axes, semi-axis lengths, focal distance, eccentricity, vertices, area, tangents, shifted equations and common mistakes. All locations and lengths are invented mathematical constructions. They are not surveyed Bukit Timah gardens, paths, buildings, MRT entrances or classroom measurements. The lesson extends coordinate and quadratic Mathematics; not all its analytic geometry is necessarily part of any student’s current examination syllabus.

Quick answer: The standard horizontal ellipse x²/a²+y²/b²=1, where a>b>0, has centre (0,0), semi-major axis a, semi-minor axis b, and foci (±c,0), where c²=a²−b². For x²/25+y²/16=1, a=5, b=4 and c=3. Every point on the ellipse satisfies PF₁+PF₂=2a=10.

Article contents

Original 2026 Mathematics study illustration with graphing materials accompanying fictional ellipse coordinates and two-focus locus constructions.

The difference between a circle and an ellipse

A circle has one centre and a fixed individual distance from that centre to every point on its circumference. An ellipse instead has two foci and a fixed sum of two distances.

A circle is a special ellipse whose foci coincide at the centre. But an elongated ellipse does not have one common radius from its centre.

This distinction explains why a circle equation and an ellipse equation have related yet different coordinate forms.

Bukit Timah is the learning setting, not a survey

A family near Beauty World, Sixth Avenue or King Albert Park may picture a symmetrical garden or an oval drawing.

The figures in this guide use an ideal flat Cartesian plane with equal horizontal and vertical units. The points were chosen to make exact calculations possible.

No conclusion identifies the geometry of an actual property or a real travel path. The model is educational, not geographical.

An ellipse is a locus of points

A locus is a set of points satisfying a rule.

Given two fixed distinct points F₁ and F₂ and a constant 2a greater than their separation, the corresponding ellipse comprises points P where PF₁+PF₂=2a.

The condition 2a>F₁F₂ is important. If the required total equals the focal separation, the locus degenerates to the segment joining the two foci rather than a nondegenerate ellipse.

Worked example 1: identify the focal separation

Let F₁=(−3,0) and F₂=(3,0).

Their separation along the horizontal axis is 3−(−3)=6 units.

The half-separation is c=3. If the sum of distances is ten, then 2a=10 and a=5, satisfying the requirement a>c.

The ellipse centre is midway between its foci

With foci (−c,0) and (c,0), the midpoint is the origin.

The line through the foci is the major axis because the ellipse extends farther in that direction than in the perpendicular direction for a>b.

Symmetry means replacing x by −x or y by −y leaves the standard ellipse equation unchanged. This is a property of the stated curve, not just a drawing convention.

Work out the two vertices

The endpoints of the major axis lie at (−a,0) and (a,0).

For a=5, the vertices are (−5,0) and (5,0).

At P=(5,0), the distances to the foci are eight and two, totalling ten. Thus it satisfies the distance-sum rule exactly.

Worked example 2: verify a major-axis endpoint

Take P=(−5,0).

Distance to F₁=(−3,0) is two. Distance to F₂=(3,0) is eight.

Their sum is 2+8=10, the same as the required 2a. Both left and right vertices pass the defining test.

The minor axis is perpendicular to the major axis

The minor axis joins (0,−b) and (0,b) for a horizontally oriented standard ellipse.

Its complete length is 2b. The full major-axis length is 2a.

Do not confuse a and b with complete axis lengths; they are semi-axis lengths, measured from centre to the corresponding boundary.

The right triangle relating a, b and c

At the top co-vertex P=(0,b), symmetry makes the distances to the two foci equal.

Their sum is 2a, so each distance must equal a. Using the right triangle with horizontal leg c and vertical leg b gives

a²=b²+c², or c²=a²−b².

This derivation connects the focal and coordinate descriptions.

Worked example 3: find the semi-minor axis

Suppose a=5 and c=3.

Then b²=a²−c²=25−9=16, so b=4 because lengths are positive.

Thus the ellipse’s full minor axis is eight units, and its top and bottom co-vertices are (0,4) and (0,−4).

Check the co-vertex by the distance rule

For P=(0,4), each focal distance is √(3²+4²)=5.

Together they give 5+5=10.

The point is four units from the centre vertically but five units from each focus. These are different lengths and should not be substituted for one another.

The standard ellipse equation

For a horizontal ellipse centred at the origin with a>b>0,

x²/a²+y²/b²=1.

The denominators are the squares of the semi-axis lengths. The larger denominator belongs to the major-axis direction.

When a=5 and b=4, the equation becomes x²/25+y²/16=1.

Worked example 4: test a point

Take P=(3,16/5)=(3,3.2).

Calculate x²/25=9/25 and y²/16=(256/25)/16=16/25.

Their sum is 25/25=1, so P lies on the ellipse. This is an exact substitution check, not a guess from a plot.

Verify that point by focal distances

For P=(3,16/5), its distance to F₂=(3,0) is 16/5=3.2.

Its distance to F₁=(−3,0) is √[6²+(16/5)²]=√[36+256/25]=√(1156/25)=34/5=6.8.

The sum is 16/5+34/5=50/5=10, confirming the original ellipse definition.

Why a point can be on the curve without being a vertex

The point (3,3.2) satisfies the ellipse equation but does not have x=±5 or y=±4.

It is an ordinary boundary point. The vertices and co-vertices are special extrema along the principal axes, not the only points on the curve.

Infinitely many other points satisfy the equation, including irrational-coordinate points.

Worked example 5: find y when x is three

Start from x²/25+y²/16=1 and set x=3.

Then y²/16=1−9/25=16/25. Multiply by sixteen to obtain y²=256/25.

Thus y=±16/5. The ellipse has both an upper and a lower point at that horizontal coordinate.

The inside region is an inequality

The ellipse boundary satisfies x²/25+y²/16=1.

The filled ellipse, including its interior, satisfies x²/25+y²/16≤1. Points with expression greater than one lie outside.

This is a coordinate feasibility region. It should not be confused with a real building boundary, public access zone or school catchment.

Worked example 6: classify points

For P=(0,0), the expression is zero, so it lies inside.

For Q=(5,0), the expression is one, so it lies on the ellipse. For R=(6,0), it is 36/25>1, so the point lies outside.

Checking squared ratios is faster than attempting to judge exact membership from an imprecise sketch.

Eccentricity describes the ellipse’s elongation

Define e=c/a for a nondegenerate ellipse, with 0≤e<1.

Since c²=a²−b², eccentricity also satisfies e²=1−b²/a².

A circle has e=0. As e approaches one from below, the ellipse becomes more elongated relative to its major axis. Eccentricity is dimensionless.

Worked example 7: calculate eccentricity

For a=5, b=4, c=3, eccentricity is e=3/5=0.6.

Its square is 0.36. The alternative formula gives 1−16/25=9/25=0.36.

Both agree. Do not report e=3 merely because the focal coordinate is three.

An ellipse with equal semi-axes is a circle

If a=b, then c²=a²−b²=0, so the two foci coincide at the centre.

The equation becomes x²/a²+y²/a²=1, or x²+y²=a².

That is a circle of radius a. Thus a circle can be viewed as the zero-eccentricity limiting case of the ellipse family.

Worked example 8: the circle special case

Take x²/25+y²/25=1.

Multiply by 25 to get x²+y²=25. The radius is five and the centre is (0,0).

Here c=0 and e=0. The major/minor labels become non-unique because every diameter has the same length.

A vertical ellipse changes the larger denominator’s direction

For a vertically elongated ellipse, write x²/b²+y²/a²=1, where a>b>0.

Its foci are (0,±c), with c²=a²−b². The major-axis vertices are (0,±a).

Students should identify the larger denominator rather than assuming the x-axis must always be major.

Worked example 9: a vertical ellipse

Consider x²/16+y²/25=1.

The major-axis direction is vertical. Centre is (0,0), a=5, b=4 and c=3.

Foci are (0,−3) and (0,3). Major vertices are (0,±5). It is a rotated counterpart of our horizontal example.

Translation moves the ellipse without changing its shape

For a horizontal ellipse centred at (h,k), the equation is

(x−h)²/a²+(y−k)²/b²=1.

Its foci are (h±c,k), vertices (h±a,k), and co-vertices (h,k±b).

Changing h,k translates every point by the same displacement. The semi-axes and eccentricity remain unchanged.

Worked example 10: a shifted ellipse

Take (x−2)²/25+(y+1)²/16=1.

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

Foci are (−1,−1) and (5,−1). Top and bottom co-vertices are (2,3) and (2,−5).

The signs inside brackets are opposite the centre coordinates

In (x−2)², centre x-coordinate is +2.

In (y+1)², centre y-coordinate is −1 because y+1=y−(−1).

A common error is to report centre (−2,+1). Checking which x and y make both squared displacements zero repairs the misunderstanding.

Worked example 11: translate a point on the curve

Original ellipse x²/25+y²/16=1 has boundary point (0,4).

Shift every point two units right and one down. The transformed point is (2,3).

Substitute into the shifted equation: (2−2)²/25+(3+1)²/16=0+16/16=1. The translation check works.

The major and minor axis lengths are not interchangeable

For the horizontal ellipse x²/25+y²/16=1, full major-axis length is 10 and full minor-axis length is 8.

The numbers 25 and 16 are squared semi-axis lengths, not the lengths themselves.

A pupil who reports major axis 25 and minor axis 16 has skipped the square roots and doubling.

Worked example 12: reverse a major-axis length

Suppose an ellipse has full major-axis length 14 and full minor-axis length 10.

Then a=7 and b=5. The focal half-separation is c=√(49−25)=√24=2√6.

A horizontal centred version has equation x²/49+y²/25=1 and foci (±2√6,0).

The area depends on the product of semi-axes

An ellipse with semi-axis lengths a and b has area A=πab.

One way to understand it is to start from a unit disk of area π and scale horizontal coordinates by a and vertical coordinates by b. Area scales by the product ab.

This is an exact geometric theorem, not an estimate based on an ellipse’s bounding rectangle.

Worked example 13: area of the five-four ellipse

Take a=5,b=4.

Area is π×5×4=20π square units, approximately 62.83.

A bounding rectangle extends from x=−5 to +5 and y=−4 to +4, with area 10×8=80. The ellipse occupies less than that rectangle.

Perimeter is more complicated than area

The perimeter of a general ellipse is not simply 2π times the average semi-axis.

Exact circumference can be expressed using an elliptic integral. Various approximations, including Ramanujan formulas, are used when a numerical length is needed.

For school-level applications, the critical habit is to avoid substituting a circle circumference formula as though an elongated ellipse had one radius.

Worked example 14: why the circle formula needs care

The ellipse with a=5,b=4 has no single radius five or four.

If someone calculates 2π×5=10π, they have found the circumference of a circle of radius five, not proved the ellipse’s perimeter.

For the ellipse, a specialised perimeter approximation is needed. Its area, however, is exactly 20π.

A tangent touches at one local point

For a standard ellipse x²/a²+y²/b²=1 and a point (x₀,y₀) on it, the tangent equation is

xx₀/a²+yy₀/b²=1.

This expression can be derived by implicit differentiation or the normal vector of the level-set function. The contact point must actually satisfy the ellipse equation.

Worked example 15: tangent at a non-axis point

Use x²/25+y²/16=1 at point P=(3,16/5).

The tangent is 3x/25+[(16/5)y]/16=1, which simplifies to 3x+5y=25.

Check P: 3(3)+5(16/5)=9+16=25. The line passes through the stated contact point.

Differentiate to verify the tangent slope

Implicit differentiation of x²/25+y²/16=1 gives

(2x/25)+(2y/16)(dy/dx)=0.

Hence dy/dx=−16x/(25y) where y≠0. At P=(3,16/5), the slope is −48/80=−3/5.

The tangent 3x+5y=25 has slope −3/5 as well.

Worked example 16: the tangent at the right vertex

At vertex (5,0), the ellipse tangent is x=5, a vertical line.

The derivative formula using division by y cannot provide a finite gradient at y=0. The geometric tangent still exists.

The standard tangent equation yields 5x/25+0=1, so x=5 directly, without division by zero.

A line can intersect, touch, or miss an ellipse

Substituting a line equation into the ellipse usually produces a quadratic equation.

Two distinct real roots give two intersection points; one repeated root corresponds to tangency; no real roots mean no intersection.

A sketch helps predict the classification, but exact algebra determines the intersection count.

Worked example 17: vertical chord through x equals three

Set x=3 in x²/25+y²/16=1.

We already found y=±16/5. The intersection points are (3,16/5) and (3,−16/5).

Their vertical separation is 32/5=6.4 units, the chord length along the line x=3.

Worked example 18: a line outside the ellipse

Take vertical line x=6.

Then x²/25=36/25>1, even before adding the nonnegative term y²/16.

No real point on x=6 can satisfy the ellipse equation. The line is outside the rightmost vertex x=5.

Line y equals x intersects symmetrically

Set y=x in x²/25+y²/16=1.

This gives x²(1/25+1/16)=1, so x²(41/400)=1.

Thus x=±20/√41 and y=x. The two intersection points are symmetric about the origin, as the ellipse equation predicts.

Worked example 19: check the positive intersection

Let x=y=20/√41.

Compute x²=400/41.

Then x²/25+y²/16=(400/41)(1/25+1/16)=(400/41)(41/400)=1.

The surd coordinates satisfy the curve exactly. Rounding early is unnecessary.

Parametric form uses sine and cosine

The standard ellipse can be parametrised by

x=a cos t, y=b sin t for 0≤t<2π.

Substitution gives x²/a²+y²/b²=cos²t+sin²t=1.

The parameter t selects points on the curve. Unless a=b, t is not generally the physical polar angle from the centre to the point.

Worked example 20: the forty-five-degree parameter

For a=5,b=4 and t=π/4, the point is

(5√2/2, 2√2).

Its ellipse equation yields (cos²t)+(sin²t)=1. Notice y/x=(4/5)tan t, so its actual geometric direction is not 45°.

The focal reflection property has assumptions

An ideal ellipse has a geometric reflection property: a ray from one focus reflecting specularly from the smooth elliptical boundary proceeds toward the other focus.

The tangent at the contact point supports the equal-angle reflection construction.

This is a theorem about ideal geometry and reflection, not evidence of any real Bukit Timah room’s acoustics, optical design or safety.

Eccentricity is a ratio, not a distance

For our ellipse, c=3 units, a=5 units, so e=3/5.

It is dimensionless because distance units cancel.

Reporting e=3 metres or e=15 confuses a focal length with a ratio and violates the ellipse restriction 0≤e<1.

Worked example 21: another eccentricity

Suppose a=10 and c=6.

Then e=6/10=0.6, and b=√(100−36)=8.

This ellipse has the same eccentricity as the original five-four example, because both semi-axis lengths are scaled by two. Similar ellipses preserve eccentricity.

Scaling preserves the ellipse’s shape ratio

Multiply all coordinates, axes and focal distances of an ellipse by the same positive factor k.

Then a,b,c are each multiplied by k, while e=c/a stays unchanged.

Area increases by k². This connects ellipse geometry to similarity and geometric scaling.

Worked example 22: double the original ellipse

Original a=5,b=4,c=3, area=20π.

Double lengths: a=10,b=8,c=6. New area is π×10×8=80π, four times the original.

Eccentricity remains 6/10=0.6. The curve is larger but has the same proportions.

An ellipse is not every oval-looking drawing

A sketch may look oval without satisfying the fixed-sum-of-distances equation.

The word “oval” is broad informal shape language; an ellipse has a precise mathematical definition.

Measurements or an explicit function are required before treating a physical curve as exactly elliptical. Visual resemblance alone is not proof.

A comparison with parabolas and hyperbolas

An ellipse fixes the sum of distances to two foci.

A hyperbola fixes the absolute difference of distances to two foci. A parabola fixes equality between distance to one focus and perpendicular distance to a directrix.

These three conics have related coordinate algebra but distinct geometric conditions. Swapping their definitions changes the curve.

Worked example 23: identify the correct conic

“Points whose distances to (−3,0) and (3,0) add to ten” describes an ellipse.

“Points whose distance difference to two fixed foci has constant absolute value” describes a hyperbola when the geometric parameters are nondegenerate.

“Points equidistant from one focus and a line” describes a parabola. The defining condition decides the model before any equation is written.

Ellipse directrices are an advanced extension

For a centred horizontal ellipse with eccentricity e=c/a>0, its directrices are vertical lines x=±a/e.

A point P on the ellipse satisfies a focus-to-corresponding-directrix distance ratio e under the standard conic convention.

This ratio formulation complements the two-focus distance-sum rule. For a circle e=0, the usual finite directrices are not defined.

Worked example 24: locate the directrices

For a=5,e=3/5, a/e=5/(3/5)=25/3.

Thus the directrices are x=−25/3 and x=25/3.

They lie beyond the major vertices at x=±5. The lines are geometric references, not physical barriers.

A focus can lie inside the filled ellipse

For a=5,b=4,c=3, the foci are at x=±3 on the major axis.

Since 3²/25=9/25<1, each focus lies strictly inside the filled ellipse.

The foci are not generally on the ellipse boundary. Confusing a focus with a vertex shifts every distance calculation.

Worked example 25: test a focus as a curve point

Take F₂=(3,0). The ellipse expression is 9/25+0=0.36, not one.

Thus F₂ lies inside, not on, the ellipse.

At the right vertex (5,0), the expression equals one. The two positions have different roles.

The sum of focal distances always equals 2a

The fixed distance sum does not depend on which boundary point we choose.

At (5,0), the focal distances are eight and two. At (0,4), they are five and five. At (3,3.2), they are 6.8 and 3.2.

All sums are ten. These examples illustrate the theorem; the geometric definition establishes it for every curve point.

The ellipse and its filled disk are different objects

The equality x²/25+y²/16=1 describes the boundary curve.

The inequality ≤1 describes the filled elliptical region.

The first is a one-dimensional boundary in the plane; the second has positive area. If a problem requests area enclosed, it refers to the filled region rather than only the curve line.

Common error 1: use c squared equals a squared plus b squared

For a horizontal ellipse, c²=a²−b², not a²+b².

The plus sign appears in the corresponding horizontal hyperbola focal formula, so the two conic families are easy to confuse.

Derive ellipse geometry from its co-vertex-to-focus right triangle to remember why a²=b²+c².

Common error 2: read the denominator as an axis length

In x²/25+y²/16=1, semi-major length is √25=5, not 25.

The complete major-axis length is 10. The semi-minor length is four and its full axis length eight.

Write a²=25 and b²=16 before taking square roots.

Common error 3: say the foci are the vertices

The vertices are at x=±5. The foci are at x=±3.

Both lie along the major axis, but they define different properties.

A quick check is that the focal separation is six, while the full major-axis length is ten.

Common error 4: call every oval an ellipse

An attractive drawing can be approximately oval yet fail the exact focal-distance rule.

A mathematical ellipse must satisfy its defining locus or equivalent equation.

For real physical shapes, the claim of exact ellipticity requires measurement or a specified ideal design, not visual appearance.

Common error 5: confuse the area with circumference

Area is πab, giving 20π square units in our example.

The circumference is a length, requires a different calculation, and is not πab or simply 2πa.

Units and definitions expose the mismatch immediately.

Common error 6: mishandle translated signs

For (x−2)²/25+(y+1)²/16=1, centre is (2,−1).

Reading the signs naively gives an incorrect centre and hence incorrect focal coordinates.

Set each bracket equal to zero to recover the shift.

Common error 7: ignore a vertical tangent

At (5,0), the tangent is x=5, although dy/dx has an undefined finite value.

A pupil who concludes “there is no tangent” has confused nonexistence of a finite slope with nonexistence of the geometric tangent.

The implicit tangent equation handles vertical and horizontal cases directly.

Primary preparation: two distances and symmetry

Younger learners can use string and two fixed pins to trace a curve, then compare its widest and narrowest spans.

Formal conic notation is not required to discover that every traced point shares a fixed total string length.

The early skill is distinguishing a single-centre circle from a two-focus construction.

Lower Secondary: coordinates and Pythagoras

Students can compute distances between coordinate points, work with squared expressions and recognise symmetry about horizontal or vertical axes.

A well-chosen three-four-five right triangle makes the focal relation understandable.

Completing the square, gradients and curve tangents should follow when those algebraic prerequisites are secure.

Advanced geometry and syllabus boundaries

Ellipses and general conic equations are higher-level analytic geometry enrichment in some learning pathways.

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

This guide does not claim that focus-directrix derivations or ellipse tangents are compulsory for every G3 student.

A twenty-minute family diagnostic

Minutes 1–5: Identify foci (−3,0),(3,0), and required distance sum ten. Find a=5,c=3.

Minutes 6–10: Apply b²=a²−c² to find b=4 and write x²/25+y²/16=1.

Minutes 11–15: Test (0,4) and (3,16/5) using both the coordinate equation and focal distances.

Minutes 16–20: State major axis length ten, minor axis length eight, area 20π, and explain why the right vertex is not the right focus.

A seven-day learning route

Day one distinguishes circle and ellipse loci. Day two uses midpoint and distance formulas.

Day three derives the relation among a,b,c. Day four identifies axes and eccentricity.

Day five practises translated equations. Day six tests tangents and intersections. Day seven checks an unfamiliar ellipse using both algebraic and focal-distance descriptions.

Practice questions with checked answers

A. Foci (±3,0), distance sum ten: a? Answer: 5.

B. With a=5,c=3, find b. Answer: 4.

C. Equation of the centred horizontal ellipse? Answer: x²/25+y²/16=1.

D. Full major and minor axis lengths? Answer: 10 and 8.

E. Eccentricity? Answer: 3/5.

F. Area? Answer: 20π square units.

G. Tangent at (3,16/5)? Answer: 3x+5y=25.

H. Foci of x²/16+y²/25=1? Answer: (0,±3).

Frequently asked questions

What defines an ellipse?

A nondegenerate ellipse is the set of points whose distances from two fixed foci add to a constant greater than the distance between the foci.

Is an ellipse just a stretched circle?

It can be obtained by scaling a circle along perpendicular axes, but its defining geometry involves a constant two-focus distance sum.

What is the difference between a vertex and a focus?

Vertices are major-axis boundary endpoints. Foci are fixed interior reference points used in the locus definition.

How do you find c?

For an ellipse with a≥b, use c²=a²−b².

What is eccentricity?

The dimensionless ratio e=c/a, ranging from zero for a circle toward one for increasingly elongated ellipses.

Is the area πab?

Yes, for semi-axis lengths a and b.

Can the tangent to an ellipse be vertical?

Yes. At a horizontal ellipse’s left or right vertex, the tangent is vertical.

Do these coordinates locate Bukit Timah properties?

No. Every point and dimension is fictional coordinate geometry.

Does every Secondary student need this topic?

Not necessarily. The precise examinable scope depends on their Mathematics programme.

How can parents recognise understanding?

The learner can derive b from a,c, check a focal distance sum, identify translated axes and verify a tangent independently.

References and connected Bukit Timah Mathematics

Use the SEAB 2027 G3 Mathematics syllabus directory for school course context. The conic constructions here are original enrichment examples, not official examination questions or measured local sites.

Continue with Coordinate Circle Equations and Tangency, Parabola Focus and Directrix, Surds and Exact Distances and Coordinate Distances and Bisectors.

The immutable Clementi Secondary 1 Mathematics tutoring reference anchors first principles, clear examples, accurate diagnosis and independent reasoning.

The lasting habit is to state the locus, distinguish semi-axes from full axes, derive the correct focal relation, verify points in two ways and keep physical interpretations within the assumptions actually provided.

Companion Mathematics in Bukit Timah guides in the new four-article batch: Hyperbolas, Distance Differences and Asymptotes · 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.