The equation of the hyperbola whose foci are at (4,6), (4,−4) respectively and having eccentricity 2 is
A
(y−1)225/8−(x−4)275/4=1
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B
(y−1)225/4−(x−4)275/4=1
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C
(y−1)275/4−(x−4)225/4=1
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D
(y−1)275/4−(x−4)225/8=1
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Solution
The correct option is B(y−1)225/4−(x−4)275/4=1 The centre of the hyperbola is the midpoint of the line joining the two foci. So the coordinates of the centre are (4+42,6−42) i.e. (4,1) Since here, the x coordinate for foci is same, therefore transverse axis lies parallel to y− axis. The equation of hyperbola with centre at (4,1) is (y−1)2a2−(x−4)2b2=1 ∵ Distance between the foci =2ae √(4−4)2+(6+4)2=2a×2 ⇒10=4a⇒a=52 ∴b2=a2(e2−1)=254(4−1)=754 Thus the equation of the hyperbola is (y−1)225/4−(x−4)275/4=1