 # Solve the Following Questions `
Question:

Let $\theta$ be the acute angle between the tangents to the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{1}=1$ and the circle $x^{2}+y^{2}=3$ at their point of intersection in the first quadrant. Then $\tan \theta$ is equal to :

1. $\frac{5}{2 \sqrt{3}}$

2. $\frac{2}{\sqrt{3}}$

3. $\frac{4}{\sqrt{3}}$

4. 2

Correct Option: , 2

Solution:

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 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

$\mathrm{m}_{1}=-\frac{1}{3 \sqrt{3}}$

And slope of tangent to the circle at $\left(\frac{3}{2}, \frac{\sqrt{3}}{2}\right)$ is $m_{2}$ $=-\sqrt{3}$

So, if angle between both curves is $\theta$ then

$\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|=\left|\frac{-\frac{1}{3 \sqrt{3}}+\sqrt{3}}{1+\left(-\frac{1}{3 \sqrt{3}}(-\sqrt{3})\right)}\right|$

$$=\frac{2}{\sqrt{3}}$$

Option (2)