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 :

• A. $\frac{2}{\sqrt{3}}$
• B. $2$
• C. $\frac{5}{2\sqrt{3}}$
• D. $\frac{4}{\sqrt{3}}$

Answer 1

The correct option is A $\frac{2}{\sqrt{3}}$

Ellipse: $x^2+9y^2=9$
Circle : $x^2+y^2=3$
$x^2=\frac{9}{4},y^2=\frac{3}{4}\Rightarrow x=\frac{\pm 3}{2},y=\frac{\pm \sqrt{3}}{2}$
Point of intersections $=(\frac{\pm 3}{2},\frac{\pm \sqrt{3}}{2})$
$\therefore$ Point of intersection in first quadrant is $(\frac{3}{2},\frac{\sqrt{3}}{2})$
Now, tangent to ellipse is
$\frac{3}{2}x+\frac{9\sqrt{3}}{2}y=9$
$\Rightarrow m_1=\frac{-1}{3\sqrt{3}}$
and tangent to circle $\frac{3}{2}x+\frac{\sqrt{3}}{2}y=9$
$\Rightarrow m_2=-\sqrt{3}$
Thus, $\tan \theta =\left|\frac{\frac{-1}{3\sqrt{3}}+\sqrt{3}}{1+(\frac{-1}{3\sqrt{3}}\times (-\sqrt{3}))}\right|$
$\therefore \tan \theta =\frac{8}{4\sqrt{3}}=\frac{2}{\sqrt{3}}$