Question

$Ifthefocioftheellipse\frac{x^2}{16}+\frac{y^2}{b^2}=1andthehyperbola\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}coincide,writethevalueof{b}^2$

Answer 1

$\text{The given equation of hyperbola is}$

$\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$

$\Rightarrow \frac{25x^2}{144}-\frac{25y^2}{81}=1$

$\Rightarrow \frac{x^2}{\frac{144}{125}}-\frac{y^2}{\frac{81}{25}}=1$

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

$\text{This equation of the hyperbola is of the from}\frac{x^2}{a_1^2}-\frac{y^2}{b_1^2}=1,$

$where{a}_1^2=\frac{144}{25}and{b}_1^2=\frac{81}{25}$

$\text{Let,e,be the eccentricity of the hyperbola.Then,}$

$e_1=\sqrt{1+\frac{b_1^2}{a_1^2}}$

$=\sqrt{1+\frac{\frac{81}{25}}{\frac{144}{25}}}$

$=\sqrt{1+\frac{81}{144}}$

$=\sqrt{\frac{144+81}{144}}$

$=\frac{15}{12}$

$=\frac{5}{4}$

$So,thecoordinatesof‘fociare(\pm a_1{e}_1,0)i.e.,(\pm 3,0)$

$\text{It is given that the foci of the elipse coincide with the foci of the hyperbola.So,the coordinates of foci of the hyperbola are}(\pm 3,0)$

$\text{Now,the given equation of ellipse is}$

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

$\Rightarrow \frac{x^2}{4^2}+\frac{y^2}{b^2}=1$

$\text{This equation of the hyperbola is of the}form\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

$Where{a}_2^2=16and{b}_2^2=b^2$

$Let,e_{2}bethe\text{eccentricity of the given ellipse.So,the coordinates of foci are}(\pm a_2{e}_2.0)$

$\therefore a_2{e}_2=3$

$\Rightarrow 4\times e_2=3(\because a_2=4)$

$\Rightarrow e_2=\frac{3}{4}$

$\Rightarrow e_2^2=\frac{9}{16}$

$Now,$

$b_2^2=a_2^2(1-e_2^2)$

$=16(1-\frac{9}{16})$

$=16\times \frac{7}{16}$

$\Rightarrow b_2^2=7(\because b_2^2-b^2)$

$\Rightarrow b^2=7$

$Hence,b^2=7$