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. $\frac{x^2}{1+{\cos }^2\alpha }-\frac{y^2}{7+{\cos }^2\alpha }=1$
• E. $\frac{x^2}{{\cos }^2\alpha }-\frac{y^2}{5-{\cos }^2\alpha }=1$

Answer 1

The correct option is A $\frac{x^2}{{\cos }^2\alpha }-\frac{y^2}{7-{\cos }^2\alpha }=1$

Given, $9x^2+16y^2=144$

$\frac{x^2}{16}+\frac{y^2}{9}=1e=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}$

Foci of the ellipse are $(\pm ae,0)$

$(\pm 4.\frac{\sqrt{7}}{4},0)\Rightarrow (\pm \sqrt{7},0)$

So the foci of hyperbola are also $(\pm \sqrt{7},0)$

Transverse axis of hyperbola $=2a=2\cos \alpha$

$\Rightarrow a=\cos \alpha$

$ae=\sqrt{7}\cos \alpha e=\sqrt{7}\Rightarrow e=\frac{\sqrt{7}}{\cos \alpha }$

For hyperbola $e^2=1+\frac{b^2}{a^2}$

$\Rightarrow \frac{7}{{\cos }^2\alpha }=1+\frac{b^2}{{\cos }^2\alpha }\Rightarrow \frac{7}{{\cos }^2\alpha }=\frac{{\cos }^2\alpha +b^2}{{\cos }^2\alpha }\Rightarrow b^2=7-{\cos }^2\alpha$

So the equation of hyperbola is

$\frac{x^2}{{\cos }^2\alpha }-\frac{y^2}{7-{\cos }^2\alpha }=1$

Option $A$ is correct.