Question

If $\theta$ be the angle between the pair of tangents drawn from $(1,2)$ to the ellipse $3x^2+2y^2=5$, then $\tan \theta$ equals to

• A. $\frac{12}{5}$
• B. $\frac{6}{5}$
• C. $\frac{6}{\sqrt{5}}$
• D. $\frac{12}{\sqrt{5}}$

Answer 1

The correct option is D $\frac{12}{\sqrt{5}}$

We know the relation $S{S}_1=T^2$

where $S\equiv 3x^2+2y^2-5S_1\equiv 3x_1^2+2y_1^2-5T\equiv 3x{x}_1+2y{y}_1-5$

Now, $S{S}_1=T^2\Rightarrow (3x^2+2y^2-5)(3x_1^2+2y_1^2-5)=(3x{x}_1+2y{y}_1-5{)}^2\Rightarrow (3x^2+2y^2-5\left)\right(3+8-5)=(3x+4y-5{)}^2\Rightarrow 9x^2-4y^2-24xy-55+30x+40y=0$

and we know that angle between 2 lines given by $a{x}^2+2hxy+b{y}^2=0$

$\tan \theta =\frac{2\sqrt{h^2-ab}}{a+b}$

Now, For

$9x^2-4y^2-24xy-55+30x+40y=0$

$a=9;b=-4;h=-12$

Hence, $\tan \theta =\frac{2\sqrt{144-(-36)}}{9-4}=\frac{2\sqrt{144+36}}{5}=\frac{2\sqrt{180}}{5}=\frac{2\ast \sqrt{180\ast 5}}{5\sqrt{5}}=\frac{2\ast 30}{5\sqrt{5}}=\frac{12}{\sqrt{5}}$