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  • The foci of a hyperbola coincide with the foci of ellipse <img alt="\mathrm{\frac{x^2}{25}+\frac{y^2}{9}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%7Bx%5E2%7D%7B25

The foci of a hyperbola coincide with the foci of ellipse \mathrm{\frac{x^2}{25}+\frac{y^2}{9}=1}. If the eccentricity of the hyperbola is 3 , then its equation is

Option: 1

\mathrm{\frac{x^2}{4}-\frac{y^2}{12}=1}


Option: 2

\mathrm{\frac{x^2}{12}-\frac{y^2}{4}=1}


Option: 3

\mathrm{\frac{x^2}{16}-\frac{y^2}{4}=1}


Option: 4

\mathrm{\frac{x^2}{16}-\frac{y^2}{128}=\frac{1}{9}}


Answers (1)

best_answer

For the ellipse \mathrm{ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 } , the foci are \mathrm{ ( \pm a e, 0)}

. In the present case, \mathrm{a=5.}
\mathrm{ \begin{aligned} & \therefore \quad 9=25\left(1-e^2\right) \Rightarrow e^2=1-\frac{9}{25}=\frac{16}{25} \\ & \therefore \quad e=\frac{4}{5} \end{aligned} }

Hence, the foci of the ellipse are \mathrm{( \pm 4,0)}

Let the hyperbola be \mathrm{\frac{x^2}{a_1{ }^2}-\frac{y^2}{b_1{ }^2}=1}                             ...........(1)

And eccentricity  \mathrm{e_1=3} (given)

\mathrm{\therefore \quad b_1^2=a_1^2\left(e_1^2-1\right)=a_1^2(9-1)}

Or \mathrm{b_1^2=8 a_1^2}                                                                     ...........(2)

As the foci of the ellipse and the hyperbola coincide, the condition is 

\mathrm{\begin{aligned} & \qquad a_1 e_1=a e=4 \\ & \text { Or } a_1 \cdot 3=4 \text { or } a_1=\frac{4}{3} \end{aligned} }

\mathrm{\therefore \quad}  From (2), \mathrm{b_1^2=8 \cdot \frac{16}{9}=\frac{128}{9}}

The required equation of the hyperbola is [by using (1)]

\mathrm{ \frac{x^2}{16}-\frac{y^2}{128}=\frac{1}{9} }

The answer is (d)

 

Posted by

Pankaj Sanodiya

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