JEE Main 2023 — Ellipse Question with Solution

We have, equation of ellipse : $15x^2+19y^2=285$

or $\frac{x^2}{19}+\frac{y^2}{15}=1$

Let the coordinate of center of circle be $(0,0)$.

Equation of circle is $x^2+y^2=16$

Equation of tangent of ellipse is

$$\begin{array}{c}y=mx\pm \sqrt{19m^2+15}\text{ or }\\ \\ mx-y\pm \sqrt{19m^2+15}=0\end{array}$$

It is also tangent to the circle $x^2+y^2=16$

Perpendicular distance from center of circle to tangent $=4$

$$\frac{\left|0-0\pm \sqrt{19m^2+15}\right|}{\sqrt{m^2+1}}=4$$

On squaring both side, we get

$$\begin{array}{c}19m^2+15=16m^2+16\\ \\ 3m^2=1\\ \\ m^2=\frac{1}{3}\\ \\ m=\pm \frac{1}{\sqrt{3}}\end{array}$$

$$\tan \theta =\frac{1}{\sqrt{3}}\Rightarrow \theta =\frac{\pi }{6}\text{ with }X\text{-axis or }$$$\frac{\pi }{3}$ with $Y$-axis

Hence, the common tangents are inclined to the minor axis of the ellipse at an angle of $\frac{\pi }{3}$.