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 :
Correct Option: , 2
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)
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