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  • The foci of the ellipse <img alt="\mathrm{\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%7Bx%5E%7B2%7D%7D%7B16%7D&plus;%5Cfrac

The foci of the ellipse \mathrm{\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1} and the hyperbola \mathrm{\frac{x^{2}}{144}-\frac{y^{2}}{81}=\frac{1}{25}} coincide, then the value of \mathrm{b^{2}} is

Option: 1

1


Option: 2

5


Option: 3

7


Option: 4

9


Answers (1)

best_answer

For the given ellipse \mathrm{a}^{2}=16. Therefore \mathrm{e}=\sqrt{1-\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}}

\Rightarrow \mathrm{e}=\sqrt{1-\frac{\mathrm{b}^{2}}{16}}=\sqrt{\frac{16-\mathrm{b}^{2}}{4}}

So, the foci of the ellipse are \mathrm{( \pm a e, 0) i.e. \left( \pm \sqrt{16-b^{2}}, 0\right)}.

For the hyperbola \mathrm{a}^{2}=\left(\frac{12}{5}\right)^{2}, \mathrm{~b}^{2}=\left(\frac{9}{5}\right)^{2}.

The eccentricity e is given by

\mathrm{e=\sqrt{1+\frac{b^{2}}{a^{2}}}=\sqrt{1+\frac{81}{144}}=\frac{15}{12}=\frac{5}{4}}
\mathrm{\text { So, } 16-b^{2}=\left(\frac{12}{5} \times \frac{5}{4}\right)^{2}=9 \Rightarrow b^{2}=7}.

Posted by

Pankaj

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