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Question

If radii of director circles of x2a2+y2b2=1 and x2a2y2b2=1 are 2r and r respectively and ee and eh be the eccentricities of the ellipse and the hyperbola respectively then

A
2e2he2e=6
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B
e2e4e2h=6
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C
4e2he2e=6
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D
e2h2e2e=0
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Solution

The correct option is C 4e2he2e=6
Equation of director circles of ellipse and hyperbola are x2+y2=a2+b2 and x2+y2=a2b2 respectively.
So,
a2+b2=4r2(1)
a2b2=r2(2)
Solving above equations, we get a2=5r22,b2=3r22
Now,
e2e=1b2a2
e2e=13r22×25r2=135=25;
e2h=1+b2a2
e2h=1+35=85
So, from given options4e2he2e=4×8525=305=6

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