# Normal to a Parabola

## Equation of the Normal at the Point $(x_1,y_1)$ of the Parabola $y^2=4ax$

Let the equation of a parabola be $y^2=4ax$ Take any point $P(x_1,y_1)$ on the parabola. Then, the equation of the tangent at point $P$ is given by $yy_1=2a(x+x_1)$

$\text{Slope of the tangent}=\frac{2a}{y_1}$ $\therefore\text{Slope of the normal}(m)=-\frac{y_1}{2a}$ Hence the equation of the normal is $y-y_1=-\frac{y_1}{2a}(x-x_1)\text{ __(1)}$

## Equation of the Normal in Slope Form (m-form)

We have, the equation of normal to the parabola $y^2=4ax$ at the point $(x_1,y_1)$, $y-y_1=-\frac{y_1}{2a}(x-x_1)$ $\text{Slope}\:(m)=-\frac{y_1}{2a}$ $\therefore y_1=-2am$ Since the point $(x_1,y_1)$ passes through the parabola, we have $y_1^2=4ax_1$ $\therefore x_1=\frac{y_1^2}{4a}=\frac{(-2am)^2}{4a}=am^2$ Thus the equation of the normal to the parabola in slope form is $y-(-2am)=m(x-am^2)$ $\therefore y=mx-2am-am^3$ This form of the normal is also known as the m-form.

## Equation of the Normal to the Parabola in Parametric Form

Let $P(at_1^2,2at_1)$ be a point on a parabola $y^2=4ax$. Then, the equation of the tangent in parametric form is given by $x-yt_1+at_1^2=0$ $\text{Slope of tangent}=\frac{1}{t_1}$ $\therefore\text{Slope of normal}=-t_1$ Hence, the equation of the normal is $y-2at_1=-t_1(x-at_1^2)$ $y=-t_1x+2at_1+at_1^3$ $\therefore y+t_1x=2at_1+at_1^3$

### Obtain the equation of the normal to the parabola $y^2=8x$ at $(2,-4)$.

Equation of the parabola is $y^2=8x$ $\therefore 4a=8\Rightarrow a=2$ Given point is $(x_1,y_1)$$=(2,-4)$. Then the equation of the normal is $y-y_1=-\frac{y_1}{2a}(x-x_1)$ $y+4=\frac{4}{4}(x-2)$ $\therefore x-y-6=0$

### Obtain the equation of the normal to the parabola $y^2=3x$ parallel to the line $y=2x+1$.

Equation of the parabola is $y^2=3x$ $\therefore 4a=3\Rightarrow a=\frac{3}{4}$ Given line is $y=2x+1$ $\text{Its Slope}=2$ Since the required normal line is parallel to the given line, $\text{Slope of normal }(m)=2$ Hence, the equation of the normal is $y=mx-2am-am^3$ $y=2x-2×\frac{3}{4}×2-\frac{3}{4}×2^3$ $\therefore y=2x-9$

### Find the condition that the line $lx+my+n=0$ may be normal to the parabola $y^2=4ax$.

The equation of the normal to the parabola $y^2=4ax$ is $y=m’x-2am’-am’^3$ $y-m’x+(2am’+am’^3)=0\text{ __(1)}$ Given line is $lx+my+n=0\text{ __(2)}$ The equations $\text{(1)}$ and $\text{(2)}$ will represent the same line if $\frac{1}{m}=-\frac{m’}{l}=\frac{2am’+am’^3}{n}$ $\therefore m’=-\frac{l}{m}$ $\text{and, }-\frac{n}{l}=2a+am’^2$ $-\frac{n}{l}=a\left(2+\frac{l^2}{m^2}\right)$ $al(2m^2+l^2)=-m^2n$ $al(2m^2+l^2)+m^2n=0$ This is the required condition.

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