JEE Main 2023 — Ellipse Question with Solution

Given,

$15x^2+19y^2=285$

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

Now plotting the diagram with given circle of radius $4$ we get, 

Now, let the equation of tangent of ellipse be $y=mx\pm \sqrt{19m^2+15}$ $\left\{\text{as} c^2=a^2{m}^2+b^2\right\}$

If it is tangent to circle $x^2+y^2=16$ then using the formula, $c^2=r^2\left(1+m^2\right)$ we get,

$\frac{\sqrt{19m^2+15}}{\sqrt{1+m^2}}=4$

$\Rightarrow 19m^2+15=16+16m^2$

$\Rightarrow 3m^2=1$

$\Rightarrow m=\pm \frac{1}{\sqrt{3}}$

$\Rightarrow \theta ={30}^{°}$ with $x-$axis,

Hence,  Angle made by tangent with minor axis i.e., with $y$ axis is $\frac{\mathrm{\pi }}{3}$