Hyperbola 1 Question 11

12. If e1 is the eccentricity of the ellipse x216+y225=1 and e2 is the eccentricity of the hyperbola passing through the foci of the ellipse and e1e2=1, then equation of the hyperbola is

(a) x29y216=1

(b) x216y29=1

(c) x29y225=1

(d) None of these

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Answer:

Correct Answer: 12. (b)

Solution:

  1. The eccentricity of x216+y225=1 is

e1=11625=35e2=53[e1e2=1]

Foci of ellipse (0,±3)

Equation of hyperbola is x216y29=1.



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