Question

The angle between the pair of tangents from the point $(1,2)$ to the ellipse $3x^2+2y^2=5$ is

• A. ${\tan }^{-1}\left(\frac{12}{5}\right)$
• B. ${\tan }^{-1}\left(\frac{6}{\sqrt{5}}\right)$
• C. ${\tan }^{-1}\left(\frac{12}{\sqrt{5}}\right)$
• D. ${\tan }^{-1}(12\sqrt{5})$

Answer 1

Answer: (C)

The combined equation of the pair of tangents drawn from $(1,2)$ to the ellipse $3x^2+2y^2=5$ is
$\left(3x^2+2y^2-5\right)(3+8-5)=(3x+4y-5{)}^2$
[Using $SS=T^2$ ]
$\Rightarrow 9x^2-24xy-4y^2+\dots =0$
If angle between these lines is $\theta$,
then $\tan \theta =\frac{2\sqrt{h^2-ab}}{a+b}$,
where $a=9,h=-12,b=-4$
$\Rightarrow \tan \theta =\frac{12}{\sqrt{5}}$
$\Rightarrow \theta ={\tan }^{-1}\left(\frac{12}{\sqrt{5}}\right)$