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Question
If the latus rectum of an ellipse is equal to half of minor axis, then find its eccentricity.
Solution
Let the equation of an ellipse is `x^2/a^2 + y^2/b^2` = 1
Length of major axis = 2a
Length of minor axis = 2b
And the length of latus rectum = `(2b^2)/a`
We have `(2b^2)/a = (2b)/2`
⇒ b = `a/2`
Now b2 = a2(1 – e2), where e is the eccentricity
⇒ b2 = 4b2(1 – e2)
⇒ 1 = 4(1 – e2)
⇒ 1 – e2 = `1/4`
⇒ e2 = `1 - 1/4`
⇒ e2 = `3/4`
∴ e = `+- sqrt(3)/2`
So, e = `sqrt(3)/2` ......[∵ e is not (–)]
Hence, the required value of eccentricity is `sqrt(3)/2`.
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