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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%5E2%7D%7B16

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, b^2=b^2 .}

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

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

Re-arranging the equation of the hyperbola, we get

\mathrm{\left[x^2\left(\frac{12}{5}\right)^2\right]-\left[y^2\left(\frac{9}{5}\right)^2\right]=1}

Comparing this with standard equation, we have 

\mathrm{a=\frac{12}{5}, b=\frac{9}{5} \text { and the centre }(0,0)}

\mathrm{\therefore \quad e^2=1+\left(\frac{b^2}{a^2}\right) \Rightarrow e=\frac{5}{4}}

Thus the foci of the hyperbola are (± ae, 0) i.e. (± 3, 0)

Since the foci of ellipse and hyperbola coincide, therefore 

\mathrm{\sqrt{\left(16-b^2\right)}=3 \quad \Rightarrow \quad b^2=7}

 

 

 

 

Posted by

Sanket Gandhi

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