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If hyperbola   \mathrm{\frac{x^2}{b^2}-\frac{y^2}{a^2}=1}   passes through the foci of ellipse

\mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1} , then the square of eccentricity of hyperbola is

Option: 1

3


Option: 2

2


Option: 3

4


Option: 4

1


Answers (1)

As the line has intercept  \mathrm{e \& e^{\prime}} on axes, it means it

passes  through the points (e,0) and (0,\mathrm{ \ e^{\prime}})

Equation of line is   \mathrm{\frac{x}{e}+\frac{y}{e^{\prime}}=1 \Rightarrow y=e^{\prime}\left(1-\frac{x}{e}\right)}          .............(1)

As   \mathrm{\frac{e}{3} \text { and } \frac{e^{\prime}}{3}}  are eccentricities of hyperbola & its

conjugate hyperbola.

\mathrm{\therefore \quad \frac{9}{e^2}+\frac{9}{\left(e^{\prime}\right)^2}=1}                              ............(2)

Equation of circle whose centre is (0,0) and radius=a is

\mathrm{x^2+y^2=a^2}

Now, line  \mathrm{y=m x+c}  is tangent to the circle \mathrm{x^2+y^2=a^2}

\mathrm{\begin{aligned} & \text { if } c^2=a^2\left(1+m^2\right) \\ & \Rightarrow \quad\left(e^{\prime}\right)^2=a^2\left(1+\frac{e^{\prime 2}}{e^2}\right) \Rightarrow\left(e^{\prime}\right)^2=a^2\left(\frac{e^2+e^{\prime 2}}{e^2}\right) \\ & \Rightarrow \quad \frac{1}{a^2}=\frac{e^2+e^{\prime 2}}{e^{\prime 2} e^2}=\frac{1}{9}\left[\left(\frac{3}{e^{\prime}}\right)^2+\left(\frac{3}{e}\right)^2\right]=\frac{1}{9} \quad(\text { Using (2)) } \\ & \therefore a=3 \end{aligned}}

 

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Sumit Saini

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