Question

Find the angle between the tangents drawn to the parabola $y=-3x^2$ from the point $(0,2)$.

Answer 1

Parabola: $y=-3x^2$

Tangents are drawn from $(0,2)=>S{S}_1=T^2$

$=>(3x^2+y)\left(3\right(0)+2)=\left[3\right(0)+\frac{y+2}{2}{]}^2$

$=>4(6x^2+6y)=y^2+4y+4$

$=>24x^2+20y-y^2-4=0$

$=>24x^2-(y^2-10-4\sqrt{6})(y-10+4\sqrt{6})=0$

$=>(2\sqrt{6}x-y+10+4\sqrt{6})(2\sqrt{6}x+y-10+4\sqrt{6})=0$

$=>m_1=2\sqrt{6}$ and $m_2=-2\sqrt{6}$

Angle between them $={tan}^{-1}\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

$={tan}^{-1}\left|\frac{4\sqrt{6}}{1-4\left(6\right)}\right|$

$={tan}^{-1}\frac{4\sqrt{6}}{23}.$