JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Ellipse

The ellipse $\frac{x^2}{19}+\frac{y^2}{15}=1$ has its major axis along the x-axis. A tangent $y=mx\pm \sqrt{19m^2+15}$ also touches the circle $x^2+y^2=16$ when $\frac{\sqrt{19m^2+15}}{\sqrt{m^2+1}}=4$, i.e. $19m^2+15=16(m^2+1)$, giving $3m^2=1$ and $m=\pm \frac{1}{\sqrt{3}}$. The tangent makes an angle ${\tan }^{-1}\frac{1}{\sqrt{3}}=\frac{\pi }{6}$ with the major axis, so its inclination to the minor axis is $\frac{\pi }{2}-\frac{\pi }{6}=\frac{\pi }{3}$.