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Question
Find the equation of the ellipse in standard form if the latus rectum has length of 6 and foci are (±2, 0).
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Solution
Let the equation of the ellipse be
`x^2/"a"^2 + y^2/"b"^2` = 1 ...(1)
Then length of latus rectum = `(2"b"^2)/"a"`
and foci = (±ae, 0)
∴ `(2"b"^2)/"a"` = 6 ...(2)
ae = 2 ...(3)
∵ b2 = a2 (1 – e2), from (2),
`(2"a"^2(1 - "e"^2))/"a"` = 6
∴ (a2 – a2e2) = 3a
∴ a2 – 4 = 3a ....[By (3)]
∴ a2 – 3a – 4 = 0
∴ (a – 4)(a + 1) = 0
∴ a = 4 or a = – 1
But a ≠ – 1
∴ a = 4
∴ from (2), `(2"b"^2)/4` = 6
∴ b2 = 12
∴ from (1), the equation of the required ellipse is
`x^2/16 + y^2/12` = 1
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