KVPY SX SB Exam 19-Nov-2017 Question Paper

Question : 84 of 120

Marks: +1, -0

An ellipse

x^2/a^2+y^2/b^2 = 1, a > b

and the parabola

x^2 = 4(y + b)

are such that the two foci of the ellipse and the end points of the latus rectum of parabola are the vertices of a square. The eccentricity of the ellipse is

$1/√13$
$2/√13$
$1/√11$
$2/√11$

Validate

Solution:

As it is square
ae = 2

4√{(ae - 2)^2 +(1- b)^2}

b = –3

a^ 2 = 13

b = 5

a^ 2 = 29

(from realation

b^2 = a ^2 (1 -e ^2 )

ae = 2

e=2/√13

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