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Q.
The angle between the tangents drawn from the point $(1,4)$ to the parabola $y^{2}=4 x$ is
TS EAMCET 2019
Solution:
Given,$y^{2}=4 x {[\therefore a=1]}$
General equation of parabola $=y=m x+\frac{a}{m}$
[where, $m$ is the slope $]$
at $(1,4) ; 4=m+\frac{a}{m}$
$\Rightarrow 4=m+\frac{1}{m} \Rightarrow m^{2}-4 m+1=0$
Let $m_{1}$ and $m_{2}$ are the roots.
$\therefore m_{1}+m_{2}=4$ and $m_{1} m_{2}=1$
Now, $\tan \theta=\frac{\left|m_{1}-m_{2}\right|}{\left|1+m_{1} m_{2}\right|}=\frac{\sqrt{\left(m_{1}+m_{2}\right)^{2}-4 m_{1} m_{2}}}{1+m_{1} m_{2}}$
$=\frac{\sqrt{(4)^{2}-4(1)}}{1+1}=\frac{\sqrt{16-4}}{2}=\frac{\sqrt{12}}{2}=\frac{2 \sqrt{3}}{2}$
$\Rightarrow \tan \theta=\sqrt{3} \Rightarrow \tan \theta=\tan \frac{\pi}{3} \Rightarrow \theta=\frac{\pi}{3}$