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