Question

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

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

Answer 1

The correct option is C ${\tan }^{-1}\left(\frac{12}{\sqrt{5}}\right)$

Here, $S=3x^2+2y^2-5=0$
$S_1=3(1{)}^2+2(2{)}^2-5=3+8-5=6$
$[\because$ Point $(1,2)]$
$T=3x\cdot x_1+2y\cdot y_1-5$
$=3x\left(1\right)+2y\left(2\right)-5$
$=3x+4y-5$
Now, use, $S{S}_1=T^2$
$(3x^2+2y^2-5)\cdot 6=(3x+4y-5{)}^2$
$18x^2+12y^2-30=9x^2+16y^2+24xy+25-30x-40y$
$\Rightarrow 9x^2-4y^2-24xy+30x+40y-55=0$
Here, $a=9,b=-4$ and $h=-12$
$\therefore \tan \theta =\frac{2\sqrt{h^2-ab}}{a+b}=\frac{2\sqrt{144+36}}{9-4}=\frac{2\sqrt{180}}{5}$
$=\frac{6\sqrt{20}}{5}=\frac{12}{\sqrt{5}}$.