Question

The length of the transverse axis of a hyperbola is $2 \,\cos \,\alpha$ . The foci of the hyperbola are the same as that of the ellipse $9x^{2}+16y^{2}=144$ . The equation of the hyperbola is

• A. $\frac{x^{2}}{\cos^{2}\alpha}-\frac{y^{2}}{7-\cos^{2} \alpha}=1$
• B. $\frac{x^{2}}{\cos^{2}\alpha}-\frac{y^{2}}{7+\cos^{2} \alpha}=1$
• C. $\frac{x^{2}}{1+\cos^{2}\alpha}-\frac{y^{2}}{7-\cos^{2} \alpha}=1$
• D. $\frac{x^{2}}{1+\cos^{2}\alpha}-\frac{y^{2}}{7+\cos^{2} \alpha}=1$

Answer 1

The Correct Option is A

Let equation of hyperbola is
$\frac{x^{2}}{a_{1}^{2}}-\frac{y^{2}}{b_{1}^{2}} =1 $
Given, $2 a_{1}=2 \cos \alpha$
$\Rightarrow a_{1}=\cos \alpha $
Also, given equation of ellipse is
Here,$ \frac{x^{2}}{16}+\frac{y^{2}}{9}=1 $
$e =\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}$
According to the given condition, Foci of hyperbola $\left(e_{1}\right)=$ Foci of ellipse $(e)$
$\Rightarrow \pm a_{1} e_{1}=\pm a e$
$\Rightarrow \cos \alpha \cdot e_{1}=4 \cdot \frac{\sqrt{7}}{4}$
$\Rightarrow \cos \alpha \cdot e_{1}=\sqrt{7}$
$\Rightarrow \cos \alpha \sqrt{1+\frac{b_{1}^{2}}{\cos ^{2} \alpha}}=\sqrt{7}$
$\Rightarrow \cos ^{2} \alpha+b_{1}^{2}=7$
$\Rightarrow b_{1}^{2}=7-\cos ^{2} \alpha$
$\therefore$ The equation of hyperbola is
$\frac{x^{2}}{\cos ^{2} \alpha}-\frac{y^{2}}{7-\cos ^{2} \alpha}=1$