InterviewSolution
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InterviewSolution
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If `a`hyperbola passes through the foci of the ellipse `(x^2)/(25)+(y^2)/(16)=1`. Its transverse and conjugate axes coincide respectively with themajor and minor axes of the ellipse and if the product of eccentricities ofhyperbola and ellipse is 1 thenthe equation ofa. hyperbola is `(x^2)/9-(y^2)/(16)=1`b. the equation of hyperbola is `(x^2)/9-(y^2)/(25)=1`c. focus of hyperbola is (5, 0)d. focusof hyperbola is `(5sqrt(3),0)` |
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Answer» Here, equation of ellipse is, `x^2/25+y^2/16 = 1` So, if e is the essentricity of ellipse, then, `e^2 = 1-b^2/a^2 = 1-16/25 = 9/25` `e = 3/5` We know, foci of ellipse is (ae,0). So, focus of given ellipse will be `(3,0)`. Now, let equation of hyperbola is, `x^2/a^2-y^2/b^2 = 1` Now, as hyperbola passes through the foci of the given ellipse`(3,0)`, We can say that.,`9/a^2 - 0 = 1=>a^2=9` Also, we are given product of eccentricity of ellipse and hyperbola is 1. If eccentricity of hyperbola is `e_h`, then, `3/5**e_h = 1=>e_h= 5/3` As we know, `e_h^2 = 1+b^2/a^2` Putting, values of `e_h and a^2`, we get,`b^2 = 16` So, equation of hyperbola is `x^2/9-y^2/16 = 1` Now, focus of hyperbola is `(+-ae_h,0) = (+-5,0)` That means, option B and option C are correct. |
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