Co-normal Points

The line perpendicular to the tangent to the curve at the point of contact is normal to the parabola. A maximum of three normals can be drawn from a point to a parabola and their feet (points) where they meet the parabola are called the co-normal points. In real life, we use normal to determine the slope and curvature of roads and bridges.

This Story also Contains

1. What is Co-normal Points?
2. Proof of Co-normal Pointsṁṁ
3. Pair of Tangent of a Parabola
4. Derivation of Pair of Tangent of a Parabola
5. Properties of co-normal points
6. Problems Based on Co-normal Points

In this article, we will cover the concept of the Co-normal Points. 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 2022.

What is Co-normal Points?

The points on the curve at which the normals pass through a common point are called co-normal points. In general, a maximum of three normals can be drawn from a point to a parabola and their feet (points) where they meet the parabola are called the co-normal points.

Proof of Co-normal Pointsṁṁ

P (h,k) is any point on the plane

equation of normal to the parabola $y^2=4 a x$ is

$
\mathrm{y}=\mathrm{mx}-2 \mathrm{am}-\mathrm{am}^3
$
It passes through the point $\mathrm{P}(\mathrm{h}, \mathrm{k})$ then,

$
\begin{aligned}
& \mathrm{k}=\mathrm{mh}-2 \mathrm{am}-\mathrm{am}^3 \\
& \text { or, } \mathrm{am}^3+\mathrm{m}(2 \mathrm{a}-\mathrm{h})+\mathrm{k}=0
\end{aligned}
$
This cubic equation have three roots say, $m_1, m_2, m_3$

$
\begin{aligned}
& m_1+m_2+m_3=0 \\
& m_1 m_2+m_2 m_3+m_3 m_1=\frac{2 a-h}{a} \\
& m_1 m_2 m_3=-\frac{k}{a}
\end{aligned}
$

Points A, B, and C in which the three normals from P (h,k) meet the parabola are called co-normal points.

Pair of Tangent of a Parabola

The combined equation of the pair of tangents drawn from an external point P(x₁,y₁) to the parabola, say S=y²-4ax=0 is SS₁=T².

Where,

$\begin{aligned} & \mathrm{S}=\mathrm{y}^2-4 \mathrm{ax} \\ & \mathrm{S}_1=\mathrm{y}_1^2-4 \mathrm{ax}_1 \\ & \mathrm{~T}=\mathrm{yy}_1-2 \mathrm{a}\left(\mathrm{x}+\mathrm{x}_1\right) \\ & \left(\mathrm{y}^2-4 \mathrm{ax}\right)\left(\mathrm{y}_1^2-4 \mathrm{ax}_1\right)=\left(\mathrm{yy}_1-2 \mathrm{a}\left(\mathrm{x}+\mathrm{x}_1\right)\right)^2\end{aligned}$

Derivation of Pair of Tangent of a Parabola

Let (h, k) be any point on either of the tangents drawn from the point P (x₁, y₁). The equation of the line joining the point (x₁, y₁) and (h, k) is

$
\begin{aligned}
\mathrm{y}-\mathrm{y}_1 & =\frac{\mathrm{k}-\mathrm{y}_1}{\mathrm{~h}-\mathrm{x}_1}\left(\mathrm{x}-\mathrm{x}_1\right) \\
\Rightarrow \quad \mathrm{y} & =\frac{\mathrm{k}-\mathrm{y}_1}{\mathrm{~h}-\mathrm{x}_1} \mathrm{x}+\frac{\mathrm{hy}_1-\mathrm{kx}_1}{\mathrm{~h}-\mathrm{x}_1}
\end{aligned}
$
For this to be a tangent to the parabola it must be of the form $y=m x+\frac{a}{m}$ comparing this two equations, we get

$
m=\frac{k-y_1}{h-x_1} \text { and } \frac{a}{m}=\frac{h y_1-k x_1}{h-x_1}
$
Therefore, by multiplication we get

$
\begin{aligned}
\quad & a=\left(\frac{k-y_1}{h-x_1}\right)\left(\frac{h y_1-k x_1}{h-x_1}\right) \\
\Rightarrow \quad & a\left(h-x_1\right)^2=\left(k-y_1\right)\left(h y_1-k x_1\right)
\end{aligned}
$

Note:

The formula SS₁ = T² works for finding a pair of tangents to any general parabola as well.

Properties of co-normal points

1) The sum of the ordinates of the feet of conormal points is zero.

2) Circle through the conormal points passes through the origin (vertex of the parabola).

3) The Centroid of the triangle formed by co-normal points lies on the x-axis or on the axis of the parabola in general.

4) Three normals can be drawn from a point to a parabola.

5) Three normals can be drawn from a point to a parabola.

Recommended Video Based on Co-normal Points

Problems Based on Co-normal Points

Example 1: If the normals drawn to the parabola, $y^2=2 x$ pass through the point $(a, 0) a \neq 0$, then 'a' must be greater than :
[JEE MAINS 2021]
Solution: For more than 3 normals (on axis)
$x>\frac{L}{2} \quad$ (where $L$ is length of L.R.)
For $y^2=2 x$
L. $R .=2$

For $(a, 0)$

$
a>\frac{\text { L.R. }}{2} \Rightarrow a>1
$

Hence, the answer is 1

Example 2: Find the locus of a point from where a normal can be drawn to the parabola which makes an angle of 45^o in an anticlockwise direction with the +x-axis

Solution: Let point be P(h,k) and let m₁, m₂, m₃ be slopes of normals drawn from this point

Thus m₁= tan 45^o =1

Now put the value of m₁=1 in relations derived for co-normal points

$\begin{aligned} & m_1+m_2+m_3=0 \Rightarrow m_2+m_3=-1 \\ & m_1 m_2+m_2 m_3+m_3 m_1=\frac{2 a-h}{a} \Rightarrow m_2+m_2 m_3+m_3=\frac{2 a-h}{a} \\ & m_1 m_2 m_3=-\frac{k}{a} \Rightarrow m_2 m_3=-\frac{k}{a}\end{aligned}$

Solving the above three equation
$
\frac{\mathrm{h}-2 \mathrm{a}}{a}=1+\frac{k}{a} h-k=3 a
$

or $x-y-3 a=0$

Hence, the answer is x-y-3a =0

Example 3: If three distinct and real normals can be drawn to $y^2=8 \mathrm{x}$ from the point $(\mathrm{a}, 0)$, then

Solution
The equation of normal in terms of $m$ is

$
y=m x-4 m-2 m^3
$
If it passes through $(\mathrm{a}, 0)$, then

$
\begin{aligned}
& \mathrm{am}-4 \mathrm{~m}-2 \mathrm{~m}^3=0 \\
& \Rightarrow \mathrm{m}\left(\mathrm{a}-4-2 \mathrm{~m}^2\right)=0 \\
& \Rightarrow \mathrm{m}=0, \mathrm{~m}^2=\frac{\mathrm{a}-4}{2}
\end{aligned}
$
For three distinct normals,

$
\begin{aligned}
& a-4>0 \\
& a>4
\end{aligned}
$

Hence, the answer is a>4

Example 4: If the normals drawn from a point to the parabola $x^2=4 y$ cut the line $y=2$ in points whose abscissa are in A.P., then the slopes of the tangents at the three co-normal points are in

Solution: Any point on $\mathrm{x}^2=4 y_{\text {is }}\left(2 \mathrm{t}, \mathrm{t}^2\right)$
Normal at this point is $y-t^2=-\frac{1}{t}(x-2 t)$
Or $\mathrm{x}+\mathrm{ty}=2 \mathrm{t}+\mathrm{t}^3$
It is drawn from $\mathrm{A}\left(\mathrm{x}_1, \mathrm{y}_1\right)$. The condition that is

$
\mathrm{x}_1+\mathrm{ty}_1=2 \mathrm{t}+\mathrm{t}^3
$
Or $\mathrm{t}^3+\mathrm{t}\left(2-\mathrm{y}_1\right)-\mathrm{x}_1=0$
Let its roots by $\mathrm{t}_1, \mathrm{t}_2, \mathrm{t}_3$

$
\therefore \quad \mathrm{t}_1+\mathrm{t}_2+\mathrm{t}_3=0
$
The intersection of (1) with $\mathrm{y}=2$ is given by

$
\begin{aligned}
& x+2 t=2 t+t^3 \\
& t^3=x
\end{aligned}
$
The abscissa of three points of intersection are

$
\mathrm{t}_1^3, \mathrm{t}_2^3, \mathrm{t}_3^3
$

They are in A.P.

$
\Rightarrow \quad 2 \mathrm{t}_2^3=\mathrm{t}_1^3+\mathrm{t}_3^3
$
Or $3 \mathrm{t}_2^3=\mathrm{t}_1^3+\mathrm{t}_2^3+\mathrm{t}_3^2$
$A s \mathrm{t}_1+\mathrm{t}_2+\mathrm{t}_3=0 \Rightarrow\left(\mathrm{t}_1+\mathrm{t}_2+\mathrm{t}_3\right)^3=0$
Or $\mathrm{t}_1^3+\mathrm{t}_2^3+\mathrm{t}_3^3=3 \mathrm{t}_1 \mathrm{t}_2 \mathrm{t}_3$
From (3) and (4)
$3 \mathrm{t}_2^3=3 \mathrm{t}_1 \mathrm{t}_2 \mathrm{t}_3$
Or $\mathrm{t}_2^2=\mathrm{t}_1 \mathrm{t}_3$
$\frac{1}{t_1} \frac{1}{t_3}=\frac{1}{t_2^2}$
But $\frac{1}{t_1}, \frac{1}{t_2}, \frac{1}{\mathrm{t}_3}$ are slopes of tangents at three conormal points.
Therefore, they are in G.P.

Example 5: Tangents PA and PB are drawn to parabola $\mathrm{y}^2=4 \mathrm{ax}$. If the slope of the bisector of the angle PAB is $\sqrt{3}$, then the locus of the point ' P ' is

Solution: The slope of tangents will be

$
\tan \left(\frac{\pi}{3}+\theta\right), \tan \left(\frac{\pi}{3}-\theta\right)
$

$
\begin{aligned}
& \text { Let } \mathrm{A} \equiv\left(\mathrm{a}^{\mathrm{t}^2}, 2 \mathrm{a}^{\mathrm{t}_1}\right), \mathrm{B} \equiv\left(\mathrm{a}^{\mathrm{t}_2^2}, 2 \mathrm{at}_2\right) \\
& \Rightarrow \quad \mathrm{P} \equiv\left[\mathrm{at}_1 \mathrm{t}_2, \mathrm{a}\left(\mathrm{t}_1+\mathrm{t}_2\right)\right]
\end{aligned}
$
We have

$
\begin{aligned}
& \mathrm{t}_1=\cot \left(\frac{\pi}{3}+\theta\right), \mathrm{t}_2=\cot \left(\frac{\pi}{3}-\theta\right) \\
& \Rightarrow \mathrm{t}_1=\frac{\frac{1}{\sqrt{3}} \cdot \cot \theta-1}{\frac{1}{\sqrt{3}}+\cot \theta}, \mathrm{t}_2=\frac{\frac{1}{\sqrt{3}} \cdot \cot \theta+1}{\frac{1}{\sqrt{3}}-\cot \theta} \\
& \Rightarrow \quad \cot \theta=\frac{\sqrt{3}+\mathrm{t}_1}{1-\sqrt{3} \mathrm{t}_1}=\frac{\mathrm{t}_2-\sqrt{3}}{1+\sqrt{3} \mathrm{t}_2} \\
& \Rightarrow \quad \sqrt{3}+3 \mathrm{t}_2+\mathrm{t}_1+\sqrt{3} \mathrm{t}_1 \mathrm{t}_2=\mathrm{t}_2-\sqrt{3}-\sqrt{3} \mathrm{t}_1 \mathrm{t}_2+3 \mathrm{t}_1 \\
& \Rightarrow \quad 2 \sqrt{3}+2 \sqrt{3} \mathrm{t}_1 \mathrm{t}_2=2\left(\mathrm{t}_1-\mathrm{t}_2\right) \\
& \Rightarrow \quad 3\left(1+\mathrm{t}_2 \mathrm{t}_2\right)^2=\left(\mathrm{t}_1-\mathrm{t}_2\right)^2=\left(\mathrm{t}_1+\mathrm{t}_2\right)^2-4 \mathrm{t}_1 \mathrm{t}_2
\end{aligned}
$
Thus, the locus of $p$ is

$
y^2=3 x^2+3 a^2+10 a x
$

Hence, the answer is $y^2=3 x^2+3 a^2+10 a x$