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Question
Find the equation of the ellipse in the cases given below:
Length of latus rectum 4, distance between foci `4sqrt(2)`, centre (0, 0) and major axis as y-axis
Solution
Given `(2"b"^2)/"a"` = 4 and 2ae = `4sqrt(2)`
Now `(2"b"^2)/"a"` = 4
2b2 = 4a
⇒ b2 = 2a
2ae = `4sqrt(2)`
ae = `sqrt(2)`
So a2e2 = 4(2) = 8
We know b2 = a2(1 – e2)
= a2 – a2e2
⇒ 2a = a2 – 8
⇒ a2 – 2a – 8 = 0
⇒ (a – 4)(a +2) = 0
⇒ a = 4 or – 2
As a cannot be negative
a = 4
So a2 = 16 and b2 = 2(4) = 8
Also major axis is along j-axis
So equation of ellipse is `x^2/8 + y^2/16` = 1
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