JEE Main 2021 — Ellipse Question with Solution

Given, $\frac{x^2}{9}+\frac{y^2}{1}=1 ...\left(1\right)$

$x^2+y^2=3 ...\left(2\right)$

On solving equations $\left(1\right) \& \left(2\right)$, we get

$\frac{x^2}{9}+\frac{3-x^2}{1}=1$

$\Rightarrow x^2+27-9x^2=9$

$\Rightarrow 8x^2=18$

$\Rightarrow x^2=\frac{9}{4}$$\Rightarrow x=\pm \frac{3}{2}$

$\Rightarrow x=\frac{3}{2}$ 

Since, point of intersection in the first quadrant.

Put, $x=\frac{3}{2}$ in equation $\left(2\right)$, we get $y=\frac{\sqrt{3}}{2}$. 

Hence, the point of intersection of the curves $\frac{x^2}{9}+\frac{y^2}{1}=1$ and $x^2+y^2=3$ in the first quadrant is $\left(\frac{3}{2}, \frac{\sqrt{3}}{2}\right)$

Now, differentiate the equation $\left(1\right)$ on both sides with respect to $x$, we get

$y^{\text{'}}=\frac{-x}{9y}$

$\therefore$Slope of tangent to the ellipse $\frac{x^2}{9}+\frac{y^2}{1}=1$ at $\left(\frac{3}{2}, \frac{\sqrt{3}}{2}\right)$ is $m_1=-\frac{1}{3\sqrt{3}}$

Now, differentiate the equation $\left(2\right)$ on both sides with respect to $x$, we get

$y^{\text{'}}=\frac{-x}{y}$

$\therefore$Slope of tangent to the circle $x^2+y^2=3$ at $\left(\frac{3}{2}, \frac{\sqrt{3}}{2}\right)$ is $m_2=-\sqrt{3}$

So, if acute angle between both curves is $\theta$ then  $\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|=\left|\frac{-\frac{1}{3\sqrt{3}}+\sqrt{3}}{1+\left(-\frac{1}{3\sqrt{3}}\right)(-\sqrt{3})}\right|=\left(\frac{2}{\sqrt{3}}\right)$