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

Answer 1

The correct option is A

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

Step 1: Calculate the foci of hyperbola.

The given equation of ellipse is $9x^2+16y^2=144$.

Divide both sides by $144$.

$\frac{x^2}{16}+\frac{y^2}{9}=1$

Eccentricity of the ellipse is calculated as

$\begin{array}{rcl}e& =& \sqrt{1-\frac{9}{16}}\\ & =& \sqrt{\frac{7}{16}}\end{array}$

Foci of the ellipse is given as $\left(\pm ae,0\right)$.

$\left(\pm 4\times \frac{\sqrt{7}}{4},0\right)\Rightarrow \left(\pm \sqrt{7},0\right)$

Foci of the hyperbola are $\left(\pm \sqrt{7},0\right)$.

Step 2: Calculate the equation of a hyperbola.

Transverse axis $\Rightarrow 2a=2\cos \alpha$

$\begin{array}{rcl}a& =& \cos \alpha \\ ae& =& \sqrt{7}\\ \cos \alpha \times e& =& \sqrt{7}\\ e& =& \frac{\sqrt{7}}{\cos \alpha }\\ e^2& =& 1+\frac{b^2}{a^2}\\ \frac{7}{{\cos }^2\alpha }& =& 1+\frac{b^2}{{\cos }^2\alpha }\\ & =& \frac{{\cos }^2\alpha +b^2}{{\cos }^2\alpha }\\ b^2& =& 7-{\cos }^2\alpha \end{array}$

So, the equation of hyperbola is $\frac{x^2}{{\cos }^2\alpha }-\frac{y^2}{7-{\cos }^2\alpha }=1$.

Hence, option $\left(A\right)$ is correct.