Chord of Contact and Diameter of Parabola - Practice Questions & MCQ

Chord of Contact and Diameter of Parabola

Chord of Contact

S is a parabola and $\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right)$ be 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 $P$.

The equation of the chord of the parabola S=y²-4ax=0 ,  from an external point P(x₁,y₁) is$1=0$ or $y_1-2 \mathrm{a}\left(x+x_1\right)=0$

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

Diameter

The locus of the mid-points 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 $y^2=4 a x$ bisecting a system of parallel chords with slope m is $\mathbf{y = 2 a / m}$

Let $y=m x+c$ be a system of parallel chord to a parabola $y^2=4 a x$. For different value 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 \quad & \mathrm{y}^2=4 \mathrm{a}\left(\frac{\mathrm{y}-\mathrm{c}}{\mathrm{~m}}\right) \\
\therefore \quad & \mathrm{my}^2-4 \mathrm{ay}+4 \mathrm{ac}=0 \\
\therefore \quad & \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}} \\
\quad[(\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 $S: y^2-4 a x=0$, whose midpoint $\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right)$ is

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

The equation of the parabola is $y^2=4 a x$
Let $A B$ is the chord and $M$ is 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 \\
& \mathrm{y}_3^2=4 \mathrm{ax}_3
\end{aligned}
$

subtract (ii) from (i), we get

$
\begin{aligned}
& \mathrm{y}_3^2-\mathrm{y}_2^2=4 \mathrm{a}\left(\mathrm{x}_3-\mathrm{x}_2\right) \\
& \Rightarrow \quad \frac{\mathrm{y}_3-\mathrm{y}_2}{\mathrm{x}_3-\mathrm{x}_2}=\frac{4 \mathrm{a}}{\mathrm{y}_3+\mathrm{y}_2}
\end{aligned}
$
$
=\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]
$

[subtract 2ax ${ }_1$ from both side]

$
\mathrm{T}=\mathrm{S}_1
$

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