Question

Find the equation of the ellipse in standard form if the dist. between its directrix is 10 and which passes through `(-sqrt(5), 2)`.

Answer 1

Let the equation of the ellipse be

`x^2/"a"^2 + y^2/"b"^2` = 1    ...(1)

Then distance between directrices = `(2"a")/"e"` = 10

∴ a = 5e  ...(2)

Since the ellipse passes through the point `(-sqrt(5), 2)`, we get

`5/"a"^2 + 4/"b"^2` = 1

∴ 5b² + 4a² = a²b²

∴ 5a²(1 – e²) + 4a² = a² · a²(1 – e²)

Dividing by a², we get,

5 – 5e² + 4 = a²(1 – e²) = 25e²(1 – e²)  ... [By (2)]

∴ 9 – 5e² = 25e² – 25e⁴

∴ 25e⁴ – 30e² + 9 = 0

∴ (5e² – 3)² = 0

∴ 5e² – 3 = 0

∴ e² = `3/5`

∴ from (2), a² = 25e² = `25(3/5)` = 15

∴ b² = a² (1 –  e²) = `15(1 - 3/5)` = 6

∴ from (1), the equation of the required ellipse is

`x^2/15 + y^2/6` = 1.