Q. Length of the normal chord of parabola $y^{2}=4x$ which makes an angle of $\frac{\pi }{4}$ with the axis of $x$ is
NTA AbhyasNTA Abhyas 2022
Solution:
The equation of the normal of slope $m$ to the parabola $y^{2}=4ax$ is $y=mx-2am-am^{3}.$
Here $y^{2}=4x,\Rightarrow a=1$ and $m=tan\left(\frac{\pi }{4}\right)=1,$ hence normal is
$y=x-2-1$
$\Rightarrow y=x-3.$
Put, $y=x-3$ in the parabola $y^{2}=4x$ to find the point of intersection of normal and parabola.
$\Rightarrow \left(x - 3\right)^{2}=4x$
$\Rightarrow x^{2}-10x+9=0$
$\Rightarrow \left(x - 1\right)\left(x - 9\right)=0$
$\Rightarrow x=1$ or $x=9.$
Put the value of $x$ in the normal to get the values of $y.$
At $x=1,y=1-3=-2$ and at $x=9,y=9-3=6.$
Thus, the point of intersection are $\left(1 , - 2\right)$ and $\left(9 , 6\right).$
Using distance formula length of normal chord is
$\sqrt{\left(9 - 1\right)^{2} + \left(6 + 2\right)^{2}}$
$=\sqrt{64 + 64}=8\sqrt{2}.$
Here $y^{2}=4x,\Rightarrow a=1$ and $m=tan\left(\frac{\pi }{4}\right)=1,$ hence normal is
$y=x-2-1$
$\Rightarrow y=x-3.$
Put, $y=x-3$ in the parabola $y^{2}=4x$ to find the point of intersection of normal and parabola.
$\Rightarrow \left(x - 3\right)^{2}=4x$
$\Rightarrow x^{2}-10x+9=0$
$\Rightarrow \left(x - 1\right)\left(x - 9\right)=0$
$\Rightarrow x=1$ or $x=9.$
Put the value of $x$ in the normal to get the values of $y.$
At $x=1,y=1-3=-2$ and at $x=9,y=9-3=6.$
Thus, the point of intersection are $\left(1 , - 2\right)$ and $\left(9 , 6\right).$
Using distance formula length of normal chord is
$\sqrt{\left(9 - 1\right)^{2} + \left(6 + 2\right)^{2}}$
$=\sqrt{64 + 64}=8\sqrt{2}.$