If focii of x2a2−y2b2=1 coincide with the focii of x225+y29=1 and eccentricity of the hyperbola is 2, then
A
a2+b2=14
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B
There is a director circle of the hyperbola
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C
Centre of the director circle is (0,0)
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D
Length of latus rectum of the hyperbola is 12
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Solution
The correct option is C Length of latus rectum of the hyperbola is 12 For the ellipse, a=5,b=3 and e=√25−925=45 ∴ae=4 Hence, the focii are (−4,0) and (4,0)
As the focii of ellipse and hyperbola coincide For the hyperbola, ae=4,e=2 ∴a=2 b2=a2(e2−1)=4(4−1)=12 ∴b=√12