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

Answer 1

Answer: (C)

Combined equation of pair of tangents from $\left(1,2\right)$ to the ellipse $3x^2+2y^2=5$ is
$\left(3x^2+2y^2-5\right)\left(3+8-5\right)={\left(3x+2y\cdot 2-5\right)}^2$
$\left(S{S}_1=T^2\right)$
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., $h=-12$
$\therefore tan\theta =\frac{2\sqrt{144+36}}{9-4}=\frac{2\sqrt{180}}{5}$
$=\frac{2\left(6\right)\sqrt{5}}{5}=\frac{12}{\sqrt{5}}$
$\therefore \theta =ta{n}^{-1}\frac{12}{\sqrt{5}}$