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Q.

The equation of ellipse circumscribing the quadrilateral whose sides are given by x =±2 and y = ±4 and distance between foci is 46, is

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a

x232+y28=1

b

x28+y232=1

c

x216+y28−1

d

x28+y216=1

answer is B.

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Detailed Solution

Let equation be x2a2+y2b2=1 where a2=b21−e2Ellipse passes through the point (2,4).∴ 4a2+16b2=1---(i) Also distance between foci is 46 . ∴    be=26∴    4b2−24+16b2=1⇒    b4−44b2+16×24=0⇒    b2−32b2−12=0⇒    b2=32( as b>4)⇒    a2=8
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