Discussion Forum

The foci of 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 value of b² is 

Answer:

Explanation:

Given Ellipse   $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1$ 

Now $b^{2}=a^{2}(1-e^{2})$

 $\Rightarrow b^{2}=16(1-e^{2}),\Rightarrow\frac{b^{2}}{16}=1-e^{2}$

$\Rightarrow e^{2}=1-\frac{b^{2}}{16}=\frac{16-b^{2}}{16}$

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

  Foci =  $(\pm ae,0)=(\pm\sqrt{16-b^{2}},0)$

 Given hyperbola : $\frac{x^{2}}{14}+\frac{y^{2}}{81}=\frac{1}{25}$ 

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

Now,   $b^{2}=a^{2}(e^{2}-1)$

$\Rightarrow (\frac{9}{5})^{2}=(\frac{12}{5})^{2}(e^{2}-1)$

 $\Rightarrow (\frac{9}{5})^{2}=(e^{2}-1)$

 $\Rightarrow e^{2}=1+\frac{81}{144}=\frac{144+81}{144}$

 $\Rightarrow e=\frac{15}{12}=\frac{5}{4}$

  Foci =  $(\pm ae,0)$= $(\pm 3,0)$

 Since foci of the given ellipse and hyperbola coincide , therefore

 $ \sqrt{16-b^{2}}=3\Rightarrow 16-b^{2}=9$

$\therefore$   b² =7