Equations of Normal to a Parabola

A Parabola is a U- shaped plane curve where any point is at an equal distance from a fixed point and from a fixed straight line. The line perpendicular to the tangent to the curve at the point of contact is normal to the parabola. In real life, we use a parabolic antenna or parabolic microphone.

In this article, we will cover the concept of the Normal of Parabola. This category falls under the broader category of Coordinate Geometry, which is a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. A total of twenty questions have been asked on JEE MAINS( 2013 to 2023) from this topic including one in 2019, three in 2021, one in 2022, and one in 2023.

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This Story also Contains

1. What is Parabola?
2. Equation of Normal in Point Form
3. Equation of Normal in Parametric Form
4. Normal in Slope Form of Parabola
5. Shifted Parabola Normal Equation
6. Point of Intersection of Normal of a Parabola
7. Normal to parabola from a point not lying on it
8. Properties of normal
9. Solved Examples Based on Normal to Parabola

What is Parabola?

A parabola is the locus of a point moving in a plane such that its distance from a fixed point (focus) is equal to its distance from a fixed line (directrix).

$\begin{equation}
\text { Hence it is a conic section with eccentricity e }=1 \text {. }
\end{equation}$

$\begin{aligned} & \frac{P S}{P M}=e=1 \\ & \Rightarrow P S=P M\end{aligned}$

Standard equation of a parabola

If the directrix is parallel to the y-axis in the standard equation of a parabola is given as
$
y^2=4 a x
$
If the directrix is parallel to the $x$-axis, the standard equation of a parabola is given as

$
x^2=4 a y
$

Equation of Normal in Point Form

The equation of the Normal at the point $P\left(x_1, y_1\right)$ to a Parabola $y^2=4 a x$ is $y-y_1=-\frac{y_1}{2 a}\left(x-x_1\right)$

Derivation of Normal in point form of parabola

The equation of tangent to the parabola $y^2=4 a x$ at $\left(x_1, y_1\right)$ is $y_1=2 a(x+x$ Slope of the tangent at $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ is $\frac{2 \mathrm{a}}{\mathrm{y}_1}$ slope of normal at $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ become $-\frac{\mathrm{y}_1}{2 \mathrm{a}}$
$\therefore$ Equation of normal at $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ is

$
\mathrm{y}-\mathrm{y}_1=-\frac{\mathrm{y}_1}{2 \mathrm{a}}\left(\mathrm{x}-\mathrm{x}_1\right)$

$
\begin{array}{c|c}
\text{Equation of Parabola} & \text{Normal at } P\left(x_1, y_1\right) \\
\hline
y^2 = 4ax & y - y_1 = -\frac{y_1}{2a} (x - x_1) \\
y^2 = -4ax & y - y_1 = \frac{y_1}{2a} (x - x_1) \\
x^2 = 4ay & x - x_1 = -\frac{x_1}{2a} (y - y_1) \\
x^2 = -4ay & x - x_1 = \frac{x_1}{2a} (y - y_1) \\
\end{array}$

Equation of Normal in Parametric Form

The equation of normal to the parabola $y^2=4 a x$ at the point $\left(\mathrm{at}^2, 2 a t\right)$ is $y+t x=2 a t+a t^3$

Derivation of Normal in Parametric Form of Parabola

The equation of the Normal at the point $\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right)$ to a Parabola $\mathrm{y}^2=4 \mathrm{ax}$ is

$
\begin{aligned}
& y-y_1=-\frac{y_1}{2 a}\left(x-x_1\right) \\
& \text { replace } x_1 \rightarrow a t^2, y_1 \rightarrow 2 a t \\
& y-2 a t=-t\left(x-a t^2\right) \Rightarrow y+t x=2 a t+a t^3
\end{aligned}
$

$
\begin{array}{l||ll}
\text{Equation of Parabola} & \text{Coordinate} & \text{Tangent Equation} \\
\hline \hline
y^2 = 4ax & \left(at^2, 2at\right) & y + t x = 2at + a t^3 \\
y^2 = -4ax & \left(-at^2, 2at\right) & y - t x = 2at + a t^3 \\
x^2 = 4ay & \left(2at, at^2\right) & x + t y = 2at + a t^3 \\
x^2 = -4ay & \left(2at, -at^2\right) & x - t y = 2at + a t^3 \\
\hline
\end{array}
$

Normal in Slope Form of Parabola

The equation of normal of parabola in slope form is given by $y=m x-2 a m-a m^3$

Derivation of Normal in Slope Form of Parabola

The equation of the Normal at the point $P\left(x_1, y_1\right)$ to a Parabola $y^2=4 a x$ is

$
y-y_1=-\frac{y_1}{2 a}\left(x-x_1\right)
$

$m$ is the slope of the tangent, then

$
\mathrm{m}=-\frac{\mathrm{y}_1}{2 \mathrm{a}} \Rightarrow \mathrm{y}_1=-2 \mathrm{am}
$

$\left(\mathrm{x}_1, \mathrm{y}_1\right)$ lies on the paarabola $\mathrm{y}^2=4 \mathrm{ax}$

$
\begin{aligned}
& \mathrm{y}_1^2=4 \mathrm{ax}_1 \Rightarrow(2 \mathrm{am})^2=4 \mathrm{ax}_1 \\
& \therefore \mathrm{x}_1=\mathrm{am}^2
\end{aligned}
$

put the value of $x_1$ and $y_1$ in the equation $y-y_1=-\frac{y_1}{2 a}\left(x-x_1\right)$ we get

$
y=m x-2 a m-a m^3
$

which is the equation of the normal of the parabola in slope form

TIP

If $c=-2 a m-a^3$, then $y=m x+c$ is the equation of normal of the parabola $y^2=$ 4ax

$
\begin{array}{c|cc}
\text { Equation of Parabola } & \text { Point of Contact } & \text { Normal Equation } \\
\hline
y^2=4 a x & \left(a m^2,-2 a m\right) & y=m x-2 a m-a m^3 \\
y^2=-4 a x & \left(-a m^2, 2 a m\right) & y=m x+2 a m+a m^3 \\
x^2=4 a y & \left(-\frac{2 a}{m}, \frac{a}{m^2}\right) & y=m x+2 a+\frac{a}{m^2} \\
x^2=-4 a y & \left(\frac{2 a}{m},-\frac{a}{m^2}\right) & y=m x-2 a-\frac{a}{m^2} \\
\hline
\end{array}
$

Shifted Parabola Normal Equation

$
\begin{array}{c|cc}
\text { Equation of Parabola } & \text { Point of Contact } & \text { Normal Equation } \\
\hline
(y-k)^2=4 a(x-h) & \left(h+a m^2, k-2 a m\right) & (y-k)=m(x-h)-2 a m-a m^3 \\
(y-k)^2=-4 a(x-h) & \left(h-a m^2, k+2 a m\right) & (y-k)=m(x-h)+2 a m+a m^3 \\
(x-h)^2=4 a(y-k) & \left(h-\frac{2 a}{m}, k+\frac{a}{m^2}\right) & (y-k)=m(x-h)+2 a+\frac{a}{m^2} \\
(x-h)^2=-4 a(y-k) & \left(h+\frac{2 a}{m}, k-\frac{a}{m^2}\right) & (y-k)=m(x-h)-2 a-\frac{a}{m^2} \\
\hline
\end{array}
$

Point of Intersection of Normal of a Parabola

Let the equation of parabola be $y^2=4 a x$

Two points, $P \equiv\left(a t_1^2, 2 a t_1\right)$ and $Q \equiv\left(a t_2^2, 2 a t_2\right)$ on the parabola $y^2=4 a x$.
Then, equation of Normal; at $P$ and $Q$ are

$
\begin{aligned}
& y_1=-t_1 x+2 a t_1+a t_1^3 \\
& y_2=-t_2 x+2 a t_2+a t_2^3
\end{aligned}
$

solving (i) and (ii) we get,

$
x=2 a+a\left(t_1^2+t_2^2+t_1 t_2\right), y=-a t_1 t_2\left(t_1+t_2\right)
$
If $R$ is the point of intersection of two normal then,

$
\mathrm{R} \equiv\left[2 \mathrm{a}+\mathrm{a}\left(\mathrm{t}_1^2+\mathrm{t}_2^2+\mathrm{t}_1 \mathrm{t}_2\right),-\mathrm{at}_1 \mathrm{t}_2\left(\mathrm{t}_1+\mathrm{t}_2\right)\right]
$

Normal to parabola from a point not lying on it

There are different methods to find the equation of normal depending upon the equation of parabola. If the equation of a parabola is in standard form say, y²= 4ax then we consider the standard equation of normal.

i.e., y=mx-2am-am³ (1)

Let this normal pass through the point (x₁, y₁) which is not lying on the parabola. y₁ = mx₁-2am-am³.

Solving this equation, we get values of m.

Since this equation is cubic in m, we get at least one real value of m. Hence, from any point on the plane, we can draw at least one normal to the parabola.

Properties of normal

We have the following properties of normal to the parabola.

1) Normal other than the axis of the parabola never passes through the focus.

2) In any parabola, normally at any point P on it bisects the external angle between the focal chord through P and the perpendicular from P to the directrix.

Solved Examples Based on Normal to Parabola

Example 1: If $\mathrm{P}(\mathrm{h}, \mathrm{k})$ be a point on the parabola $x=4 y^2$ which is nearest to the point $\mathrm{Q}(0,33)$, then the distance of P from the directrix of the parabola $y y^2=4(x+y)$ is equal to:
[JEE MAINS 2023]
SolutionEquation of normal of the parabola $x=4 y^2$
At a point $\mathrm{P}\left(\frac{\mathrm{t}^2}{16}, \frac{2 \mathrm{t}}{16}\right)$ is

$
y+t x=\frac{2 t}{16}+\frac{1}{16} t^3
$
Normal pass through $(Q=0,33)$ then

$
\begin{aligned}
& 33=\frac{t}{8}+\frac{t^3}{16} \\
& \Rightarrow t^3+2 t-528=0 \\
& \Rightarrow(t-8)\left(t^2+8+166\right)=0 \\
& \Rightarrow t=8
\end{aligned}
$
Point $P$ is $(4,1)$
Given parabola is $y^2=4(x+y)$

$
\begin{aligned}
& y^2-4 y=4 x \\
& (y-2)^2=4(x+1)
\end{aligned}
$

directrix is $x+1=-1$

$
x=-2
$
Distance of $P(4,1)$ from the directrix $x=-2$ is 6 .

Hence, the answer is 6

Example 2: Let the normal at the point P on the parabola $y^2=6 x$ pass through the point $(5,-8)$. If the tangent at $P$ to the parabola intersects its directrix at the point $Q$, then the ordinate of the point $Q$ is:
[JEE MAINS 2022]
Solution

Equation of normal : $y=-t x+2 a t+a t^3$

$
\because\left(a=\frac{3}{2}\right)
$
Since passing through $(5,-8)$, we get $\mathrm{t}=-2$
Co-ordinate of $\mathrm{Q}:(6,-6)$
Equation of tangent at $\mathrm{Q}: \mathrm{x}+2 \mathrm{y}+6=0$
Put $\mathrm{x}=\frac{-3}{2}$ to get $\mathrm{R}=\left(\frac{-3}{2}, \frac{-9}{4}\right)$

Example 3: A tangent and a normal are drawn at the point $P(2,-4)$ on the parabola $y^2=8 x$, which meet the directrix of the parabola at the points $A$ and $B$ respectively. If $Q(a, b)$ is a point such that AQBP is a square, then $2 \mathrm{a}+\mathrm{b}$ is equal to:
[JEE MAINS 2021]
Solution
For Point $P(2,-4), t=-1$
Equation of Tangent : $y=\frac{x}{t}+$ at $=-x-2 \cdots$ (1)
Equation of Normal : $y=-t x+2 a t+a t^3$

$
\begin{aligned}
& y=x-4-2 \\
\Rightarrow & y=x-6---(2)
\end{aligned}
$
Equation of Directrix: $x=-2$
So $A(-2,0) \& B(-2,-8)$
Since $A Q B P$ is a square, Mid - Point of $A B=$ Mid-Point of $P Q$

$
\Rightarrow \mathrm{Q} \text { is }(-6,-4)=(\mathrm{a}, \mathrm{b}) \text {. So } 2 a+b=-16
$
Hence, the answer is -16

Example 4: Consider the parabola with vertex $\left(\frac{1}{2}, \frac{3}{4}\right)$ and the directrix $y=\frac{1}{2}$. Let $P$ be the point where the parabola meets the line $x=-\frac{1}{2}$. If the normal to the parabola at P intersects the parabola again at the point Q , then $(\mathrm{PQ})^2$ is equal to:
[JEE MAINS 2021]
Solution

$
\text { clearly } a=\frac{1}{4}
$
Equation of parabola

$
\left(x-\frac{1}{2}\right)^2=4\left(\frac{1}{4}\right)\left(y-\frac{3}{4}\right)
$
For point $P$

$
\left(-\frac{1}{2}-\frac{1}{2}\right)^2=\left(q-\frac{3}{4}\right) \Rightarrow q=\frac{7}{4}
$
For $P Q$

$
2\left(x-\frac{1}{2}\right)=y^{\prime}
$
At $P, y^{\prime}=-2$
$\therefore$ slope of normal $=\frac{1}{2}$
Equation of PQ :

$
\begin{aligned}
& y-\frac{7}{4}=\frac{1}{2}\left(x+\frac{1}{2}\right) \\
& \Rightarrow y=\frac{x}{2}+\frac{1}{4}+\frac{7}{4}=\frac{x}{2}+2
\end{aligned}
$
For point Q

$
\begin{aligned}
& \left(x-\frac{1}{2}\right)^2=\left(\frac{x}{2}+2-\frac{3}{4}\right) \\
& \Rightarrow x=2 \\
& \therefore Q=(2,3) \\
& P Q^2=\left(2+\frac{1}{2}\right)^2+\left(3-\frac{7}{4}\right)^2=\frac{125}{16}
\end{aligned}
$
Hence, the answer is $\frac{125}{16}$

Example 5: If the point on the curve $y^2=6 x$, nearest to the point $\left(3, \frac{3}{2}\right)$ is $(\alpha, \beta)$, then $2(\alpha+\beta)$ is equal to $\qquad$
Solution
[JEE MAINS 2021]

$
y^2=6 x \Rightarrow y^2=4\left(\frac{3}{2}\right) x \Rightarrow a=\frac{3}{2}
$
Let Point P be $\left(\frac{3}{2} t^2, 3 t\right)$
Equation of normal at this point

$
y=-t x+3 t+\frac{3}{2} t^3
$
For the shortest distance, this normal will pass through $\left(3, \frac{3}{2}\right)$

$
\frac{3}{2}=-3 t+3 t+\frac{3}{2} t^3 \Rightarrow t^3=1 \Rightarrow t=1
$

Point P is $\left(\frac{3}{2}, 3\right) \Rightarrow \alpha=\frac{3}{2}, \beta=3$

$
\therefore 2(\alpha+\beta)=2\left(\frac{3}{2}+3\right)=9
$

Hence, the correct answer is 9.