Question

Length of the normal chord of parabola $y^2=4x$ which makes an angle of $\frac{\pi }{4}$ with the axis of $x$ is

• A. $8$
• B. $8\sqrt{2}$
• C. $4$
• D. $4\sqrt{2}$

Answer 1

Answer: (B)

The equation of the normal of slope $m$ to the parabola $y^2=4ax$ is $y=mx-2am-a{m}^3.$
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
$\sqrt{{\left(9-1\right)}^2+{\left(6+2\right)}^2}$
$=\sqrt{64+64}=8\sqrt{2}.$