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Q.

If the foci of a hyperbola are same as that of the ellipse x29+y225=1 and the eccentricity of the hyperbola is 15/8 times the eccentricity of the ellipse, then the smaller focal distance of the point (2,14325) on the hyperbola, is equal to  

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a

725+83

b

72583

c

142543

d

1425163

answer is B.

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Detailed Solution

a=3,b=5Focil=(9±be)               =(0,±4)eH=45×158=32Eqn    of  Hyp.x2A2y2B2=1B.eH=4B=83A2=809Directrix  y=±BeH=±169Sp=epm(Def  ofconic)Sp=72583Ans:     3

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