Explanation

(a)
The given ellipse is $\frac{x^{2}}{4} + \frac{y^{2}}{3} = 1\ \Rightarrow \ a = 2,\ b = \sqrt{3}$
$\therefore\ b^{2} = a^{2}\left( 1 - e^{2} \right)$
$\Rightarrow \ 3 = 4\left( 1 - e^{2} \right)\ \Rightarrow \ e = \frac{1}{2}$
$\therefore\ \ ae = 1$
Hence, the eccentricity $e_{1}$, of the required hyperbola is given by$\$
$1 = e_{1}\sin{\theta\ \ \Rightarrow \ e_{1} = cosec\ \theta}$
$\Rightarrow \ b^{2} = \sin^{2}{\theta\left( cosec^{2}\ \theta - 1 \right) = \cos^{2}\theta}$
Hence , the required hyperbola is
$\frac{x^{2}}{\sin^{2}\theta} - \frac{y^{2}}{\cos^{2}\theta} = 1$
or $x^{2}cosec^{2}\theta - y^{2}\sec^{2}{\theta = 1}$