---
title: "Derive standard equation of an ellipse"
subject: "Mathematics I"
chapter: 3
key: "semester_1/mathematics_i:ellipse-standard-equation-derive"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/ellipse-standard-equation-derive
site: BCT Engineering
---

# Derive standard equation of an ellipse

Mathematics I · Chapter 3: Plane Analytic Geometry

- Asked in: 2074 Chaitra, 5 marks; 2073 Shrawan; 2073 Chaitra; 2069 Chaitra; 2068 Shrawan; 2068 Chaitra (asked 6 times)
- Video: [Equation of Ellipse - Derivation](https://www.youtube.com/watch?v=qT0uUsQ3Vuc) — lets derive
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/ellipse-standard-equation-derive
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-3

## Question

Derive the equation of an ellipse in standard form.

## Short answer

Sum of distances to foci (±c,0) is 2a. Squaring √((x+c)²+y²)+√((x-c)²+y²)=2a gives x²/a² + y²/b²=1 with b²=a²-c².

## Answer

An ellipse is the locus of a point P that moves so that the sum of its distances from two fixed points (foci) is constant and greater than the distance between the foci. (Equivalently, the distance from a focus is e times the distance from a directrix, with 0 < e < 1.)

#### Derivation
Take the foci F₁(−c, 0) and F₂(c, 0) and let the constant sum be 2a, with a > c > 0. For P(x, y) on the locus:

PF₁ + PF₂ = 2a

√((x+c)² + y²) + √((x−c)² + y²) = 2a

Move the second root across and square:

(x+c)² + y² = 4a² − 4a√((x−c)² + y²) + (x−c)² + y²

4cx = 4a² − 4a√((x−c)² + y²)

a√((x−c)² + y²) = a² − cx

Square again:

a²x² − 2a²cx + a²c² + a²y² = a⁴ − 2a²cx + c²x²

(a² − c²)x² + a²y² = a²(a² − c²)

Since a > c, put b² = a² − c² (b > 0):

b²x² + a²y² = a²b²

Divide by a²b²:

**x²/a² + y²/b² = 1**

#### Properties
- Centre (0, 0); vertices (±a, 0) and (0, ±b).
- Major axis 2a (along the x-axis), minor axis 2b.
- Eccentricity e = c/a = √(1 − b²/a²) < 1; foci (±ae, 0).
- Directrices x = ±a/e.
- Latus rectum = 2b²/a.

```text
        y
        |  B(0,b)
     ___|___
   /    |    \
 -A-F1--O--F2-A'--x
   \____|____/
        | B'(0,-b)
```

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which condition defines an ellipse with foci (±c,0) and constant 2a?
   A. PF1 − PF2 = 2a with c > a
   B. PF1 + PF2 = 2a with a > c
   C. PF1 · PF2 = 2a
   D. PF1 = PF2 = a
   Answer: B. The sum of distances is the constant 2a, which must exceed the distance 2c between the foci. (Key point: Definition of the ellipse)

2. After the first squaring we get a·sqrt((x−c)^2 + y^2) = a^2 − cx. What is done next?
   A. Square again and collect terms
   B. Take logs
   C. Put x = 0
   D. Divide by a and stop
   Answer: A. Squaring again removes the remaining root and leads to (a^2 − c^2)x^2 + a^2y^2 = a^2(a^2 − c^2). (Key point: Elimination of roots)

3. Why do we put b^2 = a^2 − c^2?
   A. Because a < c
   B. Because b is the focal distance
   C. To make the ellipse a circle
   D. Because a > c, so a^2 − c^2 is positive
   Answer: D. For an ellipse a > c so a^2 − c^2 > 0; b is then the semi-minor axis. (Key point: Meaning of b^2)

4. Which pair is correct for the ellipse?
   A. x^2/a^2 − y^2/b^2 = 1 and e > 1
   B. x^2/a^2 + y^2/b^2 = 1 and e = sqrt(1 + b^2/a^2)
   C. x^2/a^2 + y^2/b^2 = 1 and e = sqrt(1 − b^2/a^2) < 1
   D. x^2/b^2 + y^2/a^2 = 1 and e = 1
   Answer: C. Dividing b^2x^2 + a^2y^2 = a^2b^2 by a^2b^2 gives the standard form; e = c/a = sqrt(1 − b^2/a^2) < 1. (Key point: Standard form and eccentricity)

## Related in Mathematics I

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- [Pedal equation of an ellipse](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Apedal-equation-ellipse) — Mathematics I, chapter 1
- [Ellipse through (1,6) and (2,3)](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Aellipse-through-two-points) — Mathematics I, chapter 3

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