Let theta be the acute angle between the tang | vedclass

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Question

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{\sq

Vedclass · Accepted Answer

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

Vedclass · Answer

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 $\left(\frac{3}{2}, \frac{\sqrt{3}}{2}ight)$

Now slope of tangent to the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{1}=1$ at $\left(\frac{3}{2}, \frac{\sqrt{3}}{2}ight)$ is

$\mathrm{m}_{1}=-\frac{1}{3 \sqrt{3}}$

And slope of tangent to the circle at $\left(\frac{3}{2}, \frac{\sqrt{3}}{2}ight)$ is $\mathrm{m}_{2}$

$=-\sqrt{3}$

So, if angle between both curves is $	heta$ then

$	an 	heta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}ight|=\left|\frac{-\frac{1}{3 \sqrt{3}}+\sqrt{3}}{1+\left(-\frac{1}{3 \sqrt{3}}(-\sqrt{3})ight)}ight|$$=\frac{2}{\sqrt{3}}$

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 :

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