## 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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