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Q.
The angle between the tangents drawn from the point $\left(2,6\right)$ to the parabola $y^{2}-4y-4x+8=0$ is
NTA AbhyasNTA Abhyas 2022
Solution:
Given parabola is $\left(y - 2\right)^{2}=4\left(x - 1\right)$
Equation of tangent in slope form to this parabola is
$y-2=m\left(x - 1\right)+\frac{1}{m}$
It passes through $\left(2,6\right),$ so
$4=m+\frac{1}{m}\Rightarrow m^{2}-4m+1=0$
If $m_{1}$ and $m_{2}$ are roots of this equation, then
$m_{1}+m_{2}=4,m_{1}\cdot m_{2}=1$
The angle between the tangents $=\tan ^{-1}\left(\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right)$
$
\begin{array}{l}
=\tan ^{-1}\left(\frac{\sqrt{\left(m_{1}+m_{2}\right)^{2}-4 m_{1} m_{2}}}{1+m_{1} m_{2}}\right)=\tan ^{-1}\left(\frac{\sqrt{16-4}}{2}\right)=\tan ^{-1} \sqrt{3} \\
=\frac{\pi}{3}
\end{array}
$