Thank you for reporting, we will resolve it shortly
Q.
The reflection of hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$ in the line $y=x$ will be a conic. The eccentricity of that conic will be
Conic Sections
Solution:
hypersC Any point on hyperbola can be taken as $(3 \sec \theta, 2 \tan \theta)$ reflection of point $(3 \sec \theta, 2 \tan \theta)$ in line y $\stackrel{x}{x}$ will be $(2 \tan \theta, 3 \sec \theta)$
$\therefore$ locus is $\frac{x^2}{4}-\frac{y^2}{9}=-1$
$\therefore$ eccentricity $=\sqrt{1+\frac{4}{9}}=\frac{\sqrt{13}}{3}$