Question

Write down the equation of the pair of tangents drawn to the ellipse $3x^2+2y^2=5$ from the point $(1,2)$, and prove that the angle between them is ${\tan }^{-1}\frac{12\sqrt{5}}{5}$.

Answer 1

$3x^2+2y^2=53x^2+2y^2-5=0$

Equation of pair of tangent from external point is $S{S}^{\prime }=T^2$

$(3x^2+2y^2-5)(3x_1^2+2y_1^2-5)=3x{x}_1+2y{y}_1-5(3x^2+2y^2-5)(3{\left(1\right)}^2+2{(2}^2)-5)={\left(3x\right(1)+2y(2)-5)}^2(3x^2+2y^2-5\left)\right(6)=9x^2+16y^2+25+24xy-40y-30x18x^2+12y^2-30=9x^2+16y^2+25+24xy-40y-30x9x^2-24xy-4y^2+30x+40y-55=0$

Angle between pair of straight lines that is $\tan \theta =\left|\frac{2\sqrt{h^2-ab}}{a+b}\right|$

Here $a=9,b=-4$ and $h=-12$

$\tan \theta =\left|\frac{2\sqrt{{\left(12\right)}^2-\left(9\right)(-4)}}{9+(-4)}\right|=\left|\frac{2\sqrt{144+36}}{5}\right|=\frac{2\times 6\sqrt{5}}{5}\Rightarrow \theta ={\tan }^{-1}\left(\frac{12\sqrt{5}}{5}\right)$

Hence proved.