Question

Sol. $\begin{array}{r} \begin{matrix} h = \frac{2sin\left( \frac{\theta_{1} + \theta_{2}}{2} \right)}{cos\left( \frac{\theta_{1} - \theta_{2}}{2} \right)} \\ k = \frac{2sin\left( \frac{\theta_{1} + \theta_{2}}{2} \right)}{cos\left( \frac{\theta_{1} - \theta_{2}}{2} \right)} \end{matrix}\#(2) \end{array}$

$$\Rightarrow \frac{h^{2}}{9} + \frac{k^{2}}{4} = \frac{1}{\cos^{2}\left( \frac{\theta_{1} - \theta_{2}}{2} \right)}$$

$$\text{~~}\begin{matrix} \text{Also}\text{ } - \lambda & \ = \frac{3cos\left( \frac{\pi}{2} + \frac{\theta_{1} + \theta_{2}}{2} \right)}{cos\left( \frac{\pi}{2} + \frac{\theta_{2} - \theta_{1}}{2} \right)} \\ & \ = \frac{3sin\left( \frac{\theta_{1} + \theta_{2}}{2} \right)}{sin\left( \frac{\theta_{2} - \theta_{1}}{2} \right)} \end{matrix}$$

Dividing (2) by (3) we get,

$$\begin{matrix} & \ tan\left( \frac{\theta_{2} - \theta_{1}}{2} \right) = \frac{3}{2}\frac{k}{\lambda} \\ & \frac{h^{2}}{9} + \frac{k^{2}}{4} = \sec^{2}\left( \frac{\theta_{2} - \theta_{1}}{2} \right) \\ & \frac{h^{2}}{9} + \frac{k^{2}}{4} = 1 + \frac{9}{4}\frac{k^{2}}{\lambda^{2}} \\ & \frac{x^{2}}{9} + \frac{y^{2}}{\left( 4\frac{\lambda^{2}}{\lambda^{2} - 9} \right)} = 1 \end{matrix}$$