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The foci of the ellipse \frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}= 1  and the hyperbola \frac{x^{2}}{144}-\frac{y^{2}}{81}= \frac{1}{25} coincide, then the value of b2 is

Option: 1

1


Option: 2

5


Option: 3

7


Option: 4

9


Answers (1)

As we learned

Eccentricities of Hyperbola -

\frac{1}{e_{1}^{2}}+\frac{1}{e_{2}^{2}}= 1

- wherein

e_{1}\, and \, e_{2}  are eccentricities of the hyperbola and its conjugate.

 

 

For the given ellipse, a2 = 16, b2 = b2.

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

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

Re-arranging the equation of the hyperbola, we get

\frac{x^2}{(12/5)^2}-\frac{y^2}{(9/5)^2}=1

Comparing this with standard equation, we have

a= \frac{12}{5},b= \frac{9}{5} \: and\: the\: centre\: \left ( 0,0 \right )

\therefore \: \: 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

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

 

Posted by

Kshitij

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