Mathematics - Ellipse Question with Solution | TestHub

Let Q be$(x_1, y_1)$
So equation tangent at Q will be
$x{x}_1+y{y}_1-\left(y+y_1\right)-1=0$
Comparing with$x+y=3$
$\left(x_1{y}_1\right) \equiv \left(\mathrm{1,2}\right)$
So R and Q will be
$\left(1 \pm \frac{2\sqrt{2}}{3}\cos \frac{3\pi }{4}, 2\pm \frac{2\sqrt{2}}{3}\sin \frac{3\pi }{4}\right)$
$\Rightarrow Q \left(\frac{5}{3} ,\frac{4}{3}\right) and \left(\frac{1}{3} ,\frac{8}{3}\right)$
Let Q lies on$\frac{x^2 }{a^2}+ \frac{y^2}{b^2}=1$
So tangent at P is
$\frac{5x}{3a^2}+\frac{4y}{3b^2}=1$
Comparing with$x+y=3$
$a^2=5, b^2=4 \Rightarrow e_1=\frac{1}{\sqrt{5}}$
And R lies on$\frac{x^2}{a_1^2}+\frac{y^2}{b_1^2}=1$
So tangent at$\frac{x}{{3a}_1^2}+\frac{8y}{{3b}_1^2}=3$
Comparing with$x+y=3$
$\Rightarrow a_1^2=1, b_1^2=8 \Rightarrow e_2=\sqrt{\frac{7}{8}}$
$\Rightarrow e_1^2+e_2^2=\frac{43}{40} and e_1{e}_2=\frac{\sqrt{7}}{2\sqrt{10}}$