Question

The angle between the pair of tangents drawn 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. $tan^{-1}\left({12}{\sqrt{5}}\right)$

Answer 1

Answer: (C)

Combined equation of pair of tangents from $\left(1,2\right)$ to the ellipse $3x^{2}+2y^{2}=5$ is
$\left(3x^{2}+2y^{2}-5\right)\left(3+8-5\right) = \left(3x+2y\cdot2 -5\right)^{2}$
$\quad\left(SS_{1} =T^{2}\right)$
i.e., $6\left(3x^{2}+2y^{2}-5\right)= \left(3x+4y-5\right)^{2} $
i.e., $18x^{2}+12y^{2}-30 = 9x^{2}+16y^{2}+25+24xy-30x-40y $
i.e., $9x^{2}-4y^{2} -24xy+30x+40y-55=0 $
$\Rightarrow 9x^{2}-24xy -4y^{2}+30x+40y-55 = 0 $
If $\theta$ is the angle between the pair of tangents, then
$ tan\,\theta =\frac{ 2\sqrt{h^{2}-ab}}{a+b} $
Here $a=9, b=-4 ; 2h= -24 $
i.e., $h=-12 $
$ \therefore tan\, \theta =\frac{ 2\sqrt{144+36}}{9-4} = \frac{2\sqrt{180}}{5} $
$= \frac{2\left(6\right)\sqrt{5}}{5} = \frac{12}{\sqrt{5}} $
$ \therefore \theta = tan^{-1} \frac{12}{\sqrt{5}}$