Question

Prove that the distance between a tangent to the parabola and the parallel normal is $a\csc \theta {\sec }^2\theta ,$ where, $\theta$ is the angle that either makes with the axis.

Answer 1

Equation of tangent is

$y=mx+\frac{a}{m}mx-y+\frac{a}{m}=0$

Equation of normal is

$y=mx-2am-a{m}^{3}mx-y-2am-a{m}^3=0$

Distance between parallel line is

$d=\frac{|\frac{a}{m}+2am+a{m}^3|}{\sqrt{1+m^2}}$

Given $m=\tan \theta$

$d=\frac{|\frac{a}{\tan \theta }+2a\tan \theta +a{\tan }^3\theta |}{\sqrt{1+{\tan }^2\theta }}d=\frac{a(\frac{1+{\tan }^2\theta }{\tan \theta }+\tan \theta (1+{\tan }^2\theta \left)\right)}{\sec \theta }d=a\cos \theta (\frac{{\sec }^2\theta }{\tan \theta }+\tan \theta {\sec }^2\theta )d=a\cos \theta {\sec }^2\theta (\frac{1}{\tan \theta }+\tan \theta )d=a\cos \theta {\sec }^2\theta \left(\frac{1+{\tan }^2\theta }{\tan \theta }\right)d=a\cos \theta {\sec }^4\theta \cot \theta d=a\csc \theta {\sec }^2\theta$