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Question

# Let the curves x2a2+y2b2=4 and x2a2−y2b2=1 have same foci. If the eccentricity of the given curves are e1 and e2 respectively, then

A
e2=3217
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B
e1=217
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C
e2=85
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D
e1=25
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Solution

## The correct options are C e2=√85 D e1=√25 From the equation of the hypberbola, the foci of the hyperbola lie on x− axis ∴ Ellipse will be horizontal ellipse. Here e1 is the eccentricity of ellipse and e2 is eccentricity of hyperbola. Now, x2a2+y2b2=4 ⇒x2(2a)2+y2(2b)2=1 ⇒e21=1−(2b)2(2a)2 ⇒e22=b2a2+1 ⇒e21+e22=2⋯(1) As foci of the ellipse will be same as that of hyperbola ∴2ae1=ae2⇒e2=2e1⋯(2) From (1) and (2), we get e1=√25,e2=√85

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