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KEAM 2013 Math Paper

Question : 65 of 120

Marks: +1, -0

\frac{x^2}{25}+\frac{y^2}{b^2}=1

\frac{x^2}{144}-\frac{y^2}{25}=\frac{1}{13}

b^2

Solution:

Given equation of ellipse,

\;\frac{x^{2}}{25}+\;\frac{y^{2}}{b^{2}}=1

.....(i)

and equation of hyperbola,

\;\frac{x^{2}}{144}-\;\frac{y^{2}}{25}=\;\frac{1}{13}

\Rightarrow \;\; \;\frac{x^{2}}{\left(\;\frac{144}{13}\right)}-\;\frac{y^{2}}{\left(\;\frac{25}{13}\right)}=1

Here,

A^{2}=\;\frac{144}{13}

and

B^{2}=\;\frac{25}{13}

\therefore

Eccentricity of hyperbola is,

e_{2}=\sqrt{\;\frac{A^{2}+B^{2}}{A^{2}}}=\sqrt{\;\frac{169}{13}\times\;\frac{13}{144}}=\;\frac{13}{12}

\therefore

Focii of hyperbola is

(\pm A e_{2}, 0)

=\left(\pm \;\frac{12}{\sqrt{13}} \times \;\frac{13}{12}, 0\right)=(\pm \sqrt{13}, 0)

Given, focii of the ellipse and hyperbola coincide. i.e., focii of an ellipse

(\pm a e_{1}, 0)=(\pm \sqrt{13}, 0)

\Rightarrow \;\; a e_{1}=\sqrt{13}

\Rightarrow \;\; 5 e_{1}=\sqrt{13}

\Rightarrow \;\; e_{1}=\;\frac{\sqrt{13}}{5}

Also,

e_{1}=\sqrt{\;\frac{a^{2}-b^{2}}{a^{2}}}

\Rightarrow \;\; e_{1}^{2} a^{2}=a^{2}-b^{2}

\Rightarrow \;\; b^{2}=a^{2}-e_{1}^{2} a^{2}=25-\;\frac{13}{25}\times 25

\therefore \;\; b^{2}=12

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