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  • If the foci of the ecllipse  <img alt="\mathrm{\frac{x^2}{16}+\frac{y^2}{b^2}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%7Bx%5E2%7D%7B16%7D&plus;%5Cf

If the foci of the ecllipse  \mathrm{\frac{x^2}{16}+\frac{y^2}{b^2}=1}  coinside with the foci of the 

hyperbola  \mathrm{\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}}, then \mathrm{b^2}  is equal to

Option: 1

7


Option: 2

8


Option: 3

10


Option: 4

9


Answers (1)

best_answer

(a) : For the hyperbola   \mathrm{\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}}  we have 

\mathrm{\begin{aligned} & a^2=\frac{144}{25}, b^2=\frac{81}{25} \\ & \therefore e^2=1+\frac{b^2}{a^2}=1+\frac{81}{144}=\frac{15^2}{12^2} \Rightarrow e=\frac{15}{12}=\frac{5}{4} \end{aligned}}

So, the coordinates of foci are  \mathrm{( \pm \text { ae, } 0) \text { i.e. }( \pm 3,0) \text {. }}

For the ellipse \mathrm{\frac{x^2}{16}+\frac{y^2}{b^2}=1}, we have \mathrm{ a^2=16}

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

so , the coordinates of foci are  \mathrm{\left( \pm \sqrt{16-b^2}, 0\right)}

Its is given that the ellipse and hyperbola have the same 

foci.

\mathrm{\therefore \sqrt{16-b^2}=3 \Rightarrow b^2=7 \text {. }}

Posted by

seema garhwal

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