Question

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

• A. $\frac{\sqrt{13}}{2}$
• B. $\frac{\sqrt{13}}{3}$
• C. $\frac{\sqrt{5}}{2}$
• D. $\frac{\sqrt{5}}{3}$

Answer 1

Answer: (B)

hypersC Any point on hyperbola can be taken as $(3\sec θ,2\tan θ)$ reflection of point $(3\sec θ,2\tan θ)$ in line y $\stackrel{x}{x}$ will be $(2\tan θ,3\sec θ)$
$∴$ locus is $\frac{x^2}{4}−\frac{y^2}{9}=−1$
$∴$ eccentricity $=\sqrt{1+\frac{4}{9}}=\frac{\sqrt{13}}{3}$