If the length of latusrectum of ellipse and are equal

MATHEMATICS · Single correct · CBSE · Mathematics · Conic Sections · Conic Sections

Problem

If the length of latusrectum of ellipse

E_1: 4(x+y-1)^2 + 2(x-y+3)^2 = 8

and

E_2: \frac{x^2}{p} + \frac{y^2}{p^2} = 1, (0 < p < 1)

are equal, then area of ellipse

E_2

, is (1)

\frac{\pi}{2}

(2)

\frac{\pi}{\sqrt{2}}

(3)

\frac{\pi}{2\sqrt{2}}

(4)

\frac{\pi}{4}