Question

Find the equation of the normal to the parabola ${{y}^{2}}=8x$ at the slope m = 2.

Answer 1

Hint: The normal of the parabola is perpendicular to the tangent of the parabola. In this question, we can use the equation of normal to the parabola ${{y}^{2}}=4ax$ at the slope m is $y=mx-2am-a{{m}^{3}}$ .

Complete step-by-step answer:
We know that the normal of the parabola is perpendicular to the tangent of the parabola.
The given equation of the parabola ${{y}^{2}}=8x$ is comparing with standard form the parabola ${{y}^{2}}=4ax$, we get
The value of a =2.
It is given that the value of slope m =2.
The equation of normal to the parabola ${{y}^{2}}=4ax$ at the slope m is given by $y=mx-2am-a{{m}^{3}}.....................(1)$ .
Now put the value of a and slope m in the equation (1), we get
 $y=2x-2(2)(2)-(2){{(2)}^{3}}$
\[y=2x-8-16\]
\[y=2x-24\]
Rearranging the terms, we get
\[2x-y-24=0\]
This is the required equation of the normal for the given parabola.

Note: Alternatively, The required equation of the normal to the given parabola in point form is given by $\left( y-{{y}_{1}} \right)=\dfrac{-{{y}_{1}}}{2a}\left( x-{{x}_{1}} \right)$. The point of contact is calculated by using $\left( {{x}_{1}},{{y}_{1}} \right)=\left( a{{m}^{2}},-2am \right)$.