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AP EAMCET Engineering 2014 Code A Solved Paper

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Question : 58

Total: 160

A hyperbola pases through a focus of the ellipse

=1. Its transverse and conjugate axes coincide respectively with the major and, ompraxex of the ellipse. The product of eccentricities is 1. Then the equation of the hyperbola is

x2
169
−
y2
25
=1
x2
144
−
y2
25
=1
x2
25
−
y2
9
=1
x2
144
−
y2
9
=1

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