- Asked 6/16
- 2074 Chaitra · 5 marks
- 2073 Shrawan
- 2073 Chaitra
- 2069 Chaitra
- 2068 Shrawan
- 2068 Chaitra
Derive standard equation of an ellipse
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.
Derive the equation of an ellipse in standard form.
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.
y
| B(0,b)
___|___
/ | \
-A-F1--O--F2-A'--x
\____|____/
| B'(0,-b)
✓ Answer checked by Claude Opus 5.5 (high effort)
Related questions
- Hyperbola: locus and standard equation · Mathematics I, chapter 3 · asked 6×
- Pedal equation of an ellipse · Mathematics I, chapter 1
- Ellipse through (1,6) and (2,3) · Mathematics I, chapter 3
- Dynamic equation of GVF, wide rectangular form · Hydraulics, chapter 8 · asked 2×
- Derive differential continuity equation · Applied Computational Fluid Dynamics, chapter 2
- A+B'C to standard minterm form · Digital Logic, chapter 3
- Laplace equation in polar form · Applied Mathematics, chapter 3 · asked 2×