Chord of Contact and Diameter of Parabola

Edited By Komal Miglani | Updated on Oct 07, 2024 09:07 AM IST

A chord (from the Latin chorda, meaning "bowstring") of a circle is a straight line segment whose endpoints both lie on a circular arc. If two tangents are drawn to a conic, like a circle, ellipse, or hyperbola from an external point, the secant line joining the points of tangency on the conic is the chord of contact of those points. Chord intersects the parabola on two points. In real life, we use chords to split curves into different parts.

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This Story also Contains

What is the Chord of Contact?
Chord of Contact of Parabola
Equation of Chord of Contact of Parabola
Diameter of the Parabola
Derivation of Diameter of Parabola
Equation of Chord Bisected at a given point
Derivation of Equation of Chord Bisected at a given point
Steps to find the Chord of Contact
Summary
Solved Examples Based on the Chord of Contact and Diameter of the Parabola

Chord of Contact and Diameter of Parabola

In this article, we will cover the concept of the Chord of Contact. 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, and one in 2023.

What is the Chord of Contact?

If two tangents are drawn to a conic, like a circle, ellipse, or hyperbola from an external point, the secant line joining the points of tangency on the conic is the chord of contact of those points. The chord of contact is a secant line formed by joining the two points of tangency on a conic.

Chord of Contact of Parabola

The chord of contact of the parabola is the line segment obtained by joining the points of tangency of the tangents to the parabola from an external point. If S is a parabola and P(x₁,y₁) is an external point to parabola S. A and B are the points of contact of the tangents drawn from P to parabola S. Then the chord AB is called the chord of contact of the parabola S drawn from an external point P.

Equation of Chord of Contact of Parabola

The equation of the chord of the parabola S=y²-4ax=0 , from an external point P(x₁,y₁) is $\mathbf{T}=\mathbf{0}$ or $\mathbf{y} \mathbf{y}_{\mathbf{1}}-2 \mathbf{a}\left(\mathbf{x}+\mathbf{x}_{\mathbf{1}}\right)=\mathbf{0}$

Note: The formula T = 0 works for finding a chord of contact from an external point for any general parabola as well

Diameter of the Parabola

The locus of the midpoints of a system of parallel chords to a parabola is known as the diameter of the parabola.

The equation of the diameter to the parabola y² = 4ax bisecting a system of parallel chords with slope m is y = 2a/m

Derivation of Diameter of Parabola

Let $y=m x+c$ be a system of parallel chords to a parabola $y^2=4 a x$. For different values of $c$, we get different chords.

Let $A\left(x_1, y_1\right)$ and $B\left(x_2, y_2\right)$ are extremities of any chord $A B$ and let its middle point is $\mathrm{M}(\mathrm{h}, \mathrm{k})$.
On solving equation,

$
\begin{array}{ll}
& \mathrm{y}^2=4 \mathrm{ax} \text { and } \mathrm{y}=\mathrm{mx}+\mathrm{c} \\
\therefore & \mathrm{y}^2=4 \mathrm{a}\left(\frac{\mathrm{y}-\mathrm{c}}{\mathrm{m}}\right) \\
\therefore & \mathrm{my}^2-4 \mathrm{ay}+4 \mathrm{ac}=0 \\
\therefore & \mathrm{y}_1+\mathrm{y}_2=\frac{4 \mathrm{a}}{\mathrm{m}} \quad \text { or } \quad \frac{\mathrm{y}_1+\mathrm{y}_2}{2}=\frac{2 \mathrm{a}}{\mathrm{m}} \\
& {[(\mathrm{h}, \mathrm{k}) \text { is the mid }- \text { point of } \mathrm{AB}]}
\end{array}
$
Hence, locus of $\mathrm{M}(\mathrm{h}, \mathrm{k})$ is $\mathrm{y}=\frac{2 \mathrm{a}}{\mathrm{m}}$

Equation of Chord Bisected at a given point

The equation of the chord of parabola S : y²- 4ax = 0, whose midpoint P(x₁,y₁) is

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

Derivation of Equation of Chord Bisected at a given point

The equation of the parabola is y² = 4ax

Let $A B$ be the chord and $M$ be the midpoint of chord $A B$
Let $A \equiv\left(x_2, y_2\right)$ and $B \equiv\left(x_3, y_3\right)$
Since, $A$ and $B$ lie on parabola,

$
\begin{aligned}
& \mathrm{y}_2^2=4 \mathrm{ax}_2 \\
& y_3^2=4 \mathrm{ax}_3 \\
& y_3^2-y_2^2=4 a\left(x_3-x_2\right) \\
& \Rightarrow \frac{\mathrm{y}_3-\mathrm{y}_2}{\mathrm{x}_3-\mathrm{x}_2}=\frac{4 \mathrm{a}}{\mathrm{y}_3+\mathrm{y}_2} \\
& =\frac{4 \mathrm{a}}{2 \mathrm{y}_1} \quad\left[\because \mathrm{M}\left(\mathrm{x}_1, \mathrm{y}_1\right) \text { is mid point of } \mathrm{AB}\right] \\
& \therefore \frac{\mathrm{y}_3-\mathrm{y}_2}{\mathrm{x}_3-\mathrm{x}_2}=\frac{2 \mathrm{a}}{\mathrm{y}_1}=\text { slope of } \mathrm{AB}
\end{aligned}
$

$\begin{aligned} & \text { Equation of } \mathrm{AB}\left(\mathrm{y}-\mathrm{y}_1\right)=\frac{2 \mathrm{a}}{\mathrm{y}_1}\left(\mathrm{x}-\mathrm{x}_1\right) \\ & \Rightarrow \quad \mathrm{yy}_1-\mathrm{y}_1^2=2 \mathrm{ax}-2 \mathrm{ax}_1 \\ & \Rightarrow \quad \mathrm{yy}_1-2 \mathrm{a}\left(\mathrm{x}+\mathrm{x}_1\right)=\mathrm{y}_1^2-4 \mathrm{ax}_1 \\ & \text { subtract } 2 \mathrm{ax}_1 \text { from both side] } \\ & \therefore \quad \mathrm{T}=\mathrm{S}_1\end{aligned}$

Let AB be the chord and M be the midpoint of chord AB

Note: The formula T = S₁ works for finding the equation of chord with a given mid-point for any general parabola as well.

Steps to find the Chord of Contact

Example 1: The length of the chord of the parabola $x^2=4 y$ having the equation $x-\sqrt{2} y+4 \sqrt{2}=0$ is:
[JEE MAINS 2019]
Solution: Equation of parabola $x^2=4 a y$ and chord $x-\sqrt{2} y+4 \sqrt{2}=0$
Solve these two equations

$
x^2=4\left(\frac{x+4 \sqrt{2}}{\sqrt{2}}\right)
$

We can find the equation of the chord of contact of the parabola by following the below steps:

Suppose we have a parabola with the equation y² = 4ay and ( e,f) is the external point from which both the tangents are drawn.

Step 1: First we need to find the value of a by comparing the given equation with y² = 4ay

Step 2: Next we substitute these values in the formula. $\mathbf{T}=\mathbf{0}$ or $\mathbf{y} \mathbf{y}_{\mathbf{1}}-2 \mathbf{a}\left(\mathbf{x}+\mathbf{x}_{\mathbf{1}}\right)=\mathbf{0}$

Summary

The chord of the parabola intersects the parabola at two points and divides it into two segments. The properties of the chord tell about the geometric properties of the parabola and its chord. Understanding of Parabola helps us to analyze the parabola.

Solved Examples Based on the Chord of Contact and Diameter of the Parabola

$
\begin{aligned}
& x^2=4\left(\frac{x+4 \sqrt{2}}{\sqrt{2}}\right) \\
& \sqrt{2} x^2=4 x+16 \sqrt{2} \\
& x_1+x_2=2 \sqrt{2} ; x_1 x_2=-16
\end{aligned}
$

Similarly,

$
\begin{aligned}
& (\sqrt{2} y-4 \sqrt{2})^2=4 y \\
& =>2 y^2-20 y+32=0 \\
& y_1+y_2=10 ; y_1 y_2=16
\end{aligned}
$
Length of chord $=\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}$

$
\begin{aligned}
& =\sqrt{(2 \sqrt{2})^2+64+(10)^2-4(16)} \\
& =\sqrt{108} \\
& =6 \sqrt{3}
\end{aligned}
$

Hence, the answer is $6\sqrt3$

Example 2: Let $P Q$ be a focal chord of the parabola $y^2=36 x$ of length 100 , making an acute angle with the positive $x$-axis. Let the ordinate of $P$ be positive and $M$ be the point on the line segment PQ such that $\mathrm{PM}: \mathrm{MQ}=3: 1$. Then which of the following points does NOT lie on the line passing through $M$ and perpendicular to the line PQ?
[JEE MAINS 2023]
Solution

$
\begin{aligned}
& 9\left(t+\frac{1}{t}\right)^2=100 \\
& \mathrm{t}=3
\end{aligned}
$

$\begin{aligned} & \Rightarrow \mathrm{P}(81,54) \& Q(1,-6) \\ & \mathrm{M}(21,9) \\ & \Rightarrow \mathrm{L} \text { is }(\mathrm{y}-9)=\frac{-4}{3}(x-21) \\ & 3 \mathrm{y}-27=-4 x+84 \\ & 4 \mathrm{x}+3 \mathrm{y}=111\end{aligned}$

Hence, the answer is (-3, 43)

Example 3: Tangents drawn from the point $(-8,0)$ to the parabola $y^2=8 x$ touch the parabola at $P$ and $Q$. If $F$ is the focus of the parabola, then the area of the triangle $P F Q$ (in sq. units) is equal to :

Solution: We know that,
Standard equation of parabola $-y^2=4 a x$
The equation of $C O C P Q$ is $T=0$
$
T \equiv 4\left(x+x_1\right)-y y_1=0
$
Where $\left(x_1, y_1\right)$ is $(-8,0)$
The chord of contact is $x=8$
$P(8,8)$ and $Q(8,-8)$
focus $=(2,0)$
$\triangle P Q F=\frac{1}{2}(8-2) \times(8+8)=48$ sq units.

Hence, the answer is 48 sq units

Example 4: If two distinct chords of a parabola $y^2=4 a x$, passing through $(a, 2 a)$ are bisected on the line $x+y=1$, then the length of the latus rectum can be less than
Solution: Any point on the line $x+y=1$ can be taken $(t, 1-t)$ equation of the chord, with this as the midpoint

$y(1-t)-2 a(x+t)=(1-t)^2-4 a t$ it passes through $(a, 2 a)$ So, $t^2-2 t+2 a^2-2 a+1=0$, this should have two distinct real roots so, $a^2-a<0,0<a<1$.

So, the length of the latus rectum < 4.
Hence, the answer is 4

Example 5: The point $(1,2)$ is one extremity of the focal chord of parabola $y^2=4 x$. The length of this focal chord is

Solution: The parabola $y^2=4 x$. Here $\mathrm{a}=1$ and focus is $(1,0)$.

The focal chord is ASB. This is clearly the latus rectum of the parabola, and its value = 4.

Hence the correct answer is 4