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