Question

Find the equation of the normal to the parabola ${{y}^{2}}=4x$ at the point (1, 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 point $\left( {{x}_{1}},{{y}_{1}} \right)$ is$\left( y-{{y}_{1}} \right)=\dfrac{-{{y}_{1}}}{2a}\left( x-{{x}_{1}} \right)$.

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 ${{y}^{2}}=4x$ is comparing with standard form the parabola${{y}^{2}}=4ax$, we get
The value of a =1.
The point of contact of the given parabola and the required the equation of the normal is $\left( {{x}_{1}},{{y}_{1}} \right)=\left( 1,2 \right)$.
The equation of normal to the parabola ${{y}^{2}}=4ax$ at the point $\left( {{x}_{1}},{{y}_{1}} \right)$ is given by $\left( y-{{y}_{1}} \right)=\dfrac{-{{y}_{1}}}{2a}\left( x-{{x}_{1}} \right)......................(1)$
Now put the value of a and given point in the equation (1), we get
$\left( y-2 \right)=\dfrac{-2}{2\times 1}\left( x-1 \right)$
Cancelling 2 on the right side, we get
$\left( y-2 \right)=-\left( x-1 \right)$
$y-2=-x+1$
Rearranging the terms, we get
$x+y-1-2=0$
$x+y-3=0$
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 slope form is given by $y=mx-2am-a{{m}^{3}}$. Where m is the slope of the normal to the given parabola.