Question

Sol. (a,b,c,d) For ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{7} = 1$

and hyperbola $\frac{x^{2}}{(144/25)} - \frac{y^{2}}{(81/25)} = 1$

$$16 - 7 = 16e_{1}^{2}$$

and $\frac{144}{25} + \frac{81}{25} = \frac{144}{25}e_{2}^{2}$

$$\begin{matrix} \frac{3}{4} & \ = e_{1} \\ e_{2} & \ = \frac{15}{12} = \frac{5}{4} \\ e_{2} & \ = \frac{5}{4} \\ Foci & \ \equiv ( \pm 3,0) \end{matrix}$$

For point of intersection of hyperbola and ellipse

$$\begin{matrix} & \frac{x^{2}}{144} + \frac{y^{2}}{63} = \frac{1}{9} \\ & \frac{x^{2}}{144} - \frac{y^{2}}{81} = \frac{1}{25} \\ & \ \Rightarrow y^{2}\left( \frac{1}{63} + \frac{1}{81} \right) = \frac{1}{9} - \frac{1}{25} = \frac{16}{9 \times 25} \\ & \ \Rightarrow \ y^{2}\left( \frac{9 + 7}{81 \times 7} \right) = \frac{16}{9 \times 25} \\ & \ \Rightarrow \ y = \pm \frac{3\sqrt{7}}{5} \\ & x^{2} = \left( 1 - \frac{63}{25 \times 7} \right)16 = \frac{16 \times 16}{25} \\ & x = \pm \frac{16}{5} \end{matrix}$$

Point of intersection $\equiv \left( \pm \frac{16}{5}, \pm \frac{3\sqrt{7}}{5} \right)$

Equation of asymptote $y = \pm \frac{3}{4}x$

$$\text{~}\text{Point,}\text{~}P \equiv \left( \pm \frac{16}{5}, \pm \frac{12}{5} \right)$$

Sum of square of coordinates of $P = \frac{256 + 144}{25} = 16$

$$x^{2} + y^{2} = 16 = a^{2}$$

$\Rightarrow P$ lies on auxiliary circle.