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

#### Solution

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

∴ 5b^{2} + 4a^{2} = a^{2}b^{2}

∴ 5a^{2}(1 – e^{2}) + 4a^{2} = a^{2} · a^{2}(1 – e^{2})

Dividing by a^{2}, we get,

5 – 5e^{2} + 4 = a^{2}(1 – e^{2}) = 25e^{2}(1 – e^{2}) ... [By (2)]

∴ 9 – 5e^{2} = 25e^{2} – 25e^{4}

∴ 25e^{4} – 30e^{2} + 9 = 0

∴ (5e^{2} – 3)^{2} = 0

∴ 5e^{2} – 3 = 0

∴ e^{2} = `3/5`

∴ from (2), a^{2} = 25e = `25(3/5)` = 15

∴ b^{2} = a^{2} (1 – e^{2}) = `15(1 - 3/5)` = 6

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

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