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1)

If e1  is the eccentricity  of the ellipse   x216+y225=1  and e2 is the eccentricity of a hyperbola  passing through  the foci of the given ellipse  and e1e2=1 , then the equation of such a hyperbola among the following is 


A) x29y216=1

B) y29x216=1

C) x29y225=1

D) x225y29=1

Answer:

Option B

Explanation:

Equation of ellipse    x216+y225=1 

 Foci   =(0,±3)

     e1=11625=35 given e1e2 =1

             (where e2 is eccentricity of hyperbola )

Let equation of hyperbola -x2a2+y2b2=1

 its passes through (0, ±3)

     b2=9

     e2=1+a2b2

    e22=1+a2b2

    1e22=1+a29259=1+a29

       a2=16

Hence, the equation of hyperbola  y29x216=1