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

Sum

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

∴ 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 = 25e = 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.

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