Equation of 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). It is a conic section with eccentricity $\mathrm{e}=1$. Let focus of parabola is $\mathrm{S}(\mathrm{a}, 0)$ and directrix be $\mathrm{x}+\mathrm{a}=0$. Then, the general equation of a parabola is $y^2=4 a x$. In real life, we use Parabolas in bridges, telescopes, satellites, etc.

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This Story also Contains

1. What is Parabola?
2. Solved Examples Based on the Equation of a parabola
3. Summary

In this article, we will cover the concept of the Equation of a parabola when the equation of axis and tangent at the vertex and latus rectum is given. 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-five questions have been asked on JEE MAINS( 2013 to 2023) from this topic including one in 2022, and one in 2023.

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).

Hence it is a conic section with eccentricity $\mathrm{e}=1$.

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

2nd form of Parabola

Equation of parabola when the equation of axis, tangent at the vertex, and latus rectum length are given

Let the equation of the axis be $\mathrm{I} x+\mathrm{my}+\mathrm{n}=0$ and the equation of the tangent at the vertex be $m x-l y+r=0$.

$\begin{equation}
\begin{aligned}
&\text { The equation of parabola is }\\
&\begin{gathered}
(\mathrm{PM})^2=(\text { Latusrectum }) \cdot(\mathrm{PN}) \\
\Rightarrow\left(\frac{\mathrm{lx}+\mathrm{my}+\mathrm{n}}{\sqrt{\mathrm{l}^2+\mathrm{m}^2}}\right)^2=(\text { Latusrectum }) \cdot\left(\frac{\mathrm{mx}-\mathrm{ly}+\mathrm{r}}{\sqrt{\mathrm{m}^2+\mathrm{l}^2}}\right)
\end{gathered}
\end{aligned}
\end{equation}$

Shifted Parabola

- If the parabola $y^2=4 a x$ is shifted (without any rotation) to a new position with a new vertex as (p,q), then the equation of a new parabola is $(y-q)^2=4 a(x-p)$

- If the parabola $y^2=-4 a x$ is shifted (without any rotation) to $a$ new position with a new vertex as ( $p, q$ ), then the equation of a new parabola is $(y-q)^2=-4 a(x-p)$
- If the parabola $x^2=4$ ay is shifted (without any rotation) to a new position with a new vertex as ( $p, q$ ), then the equation of a new parabola is $(x-p)^2=4 a(y-q)$
- If the parabola $x^2=-4$ ay is shifted (without any rotation) to a new position with a new vertex as ( $p, q$ ), then the equation of a new parabola is $(x-p)^2=-4 a(y-q)$

Note: Parametric point of $(y-q)^2=4 a(x-p)$ is $\left(p+a t^2, q+2 a t\right)$

Equation of a Parabola when the Vertex is $(h, k)$ and the Parabolic Curve

The equation of the parabola when the axis is parallel to the $x$-axis $
y^2=4 a x
$ can be written as $(y-0)^2=4 a(x-0)$

The vertex of the parabola is $O(0,0)$. Now the origin is shifted to $O^{\prime}(h, k)$ without changing the direction of axes, its equation becomes $
(y-k)^2=4 a(x-h)
$

The parametric equation of the curve $(y-k)^2=4 a(x-h)$ are $x-h+a t^2$ and $y=k+2 a t$

Thus its focus is $S(a+h, k)$, latus rectum $=4 a$ and the equation of the directrix is $
x=h-a \text {, i.e. } x+a-h=0
$

Another equation of the parabola when the axis is parallel to the $y$-axis is $
(x-h)^2=4 a(y-k)
$

Its focus is $\mathrm{S}(\mathrm{h}, \mathrm{a}+\mathrm{k})$, latus rectum $=4 \mathrm{a}$ and the equation of the directrix is $y=k-a$, i.e. $y+a-k=0$

Parabolic Curve

The equations of the form $\mathrm{y}=\mathrm{Ax}^2+\mathrm{Bx}+\mathrm{C}$ and $\mathrm{x}=\mathrm{Ay}^2+\mathrm{By}+\mathrm{C}$ are always represented as parabolas, generally called as parabolic curve.
Now,

$
\begin{aligned}
y & =A x^2+B x+C \\
& =A\left(x^2+\frac{B}{A} x+\frac{C}{A}\right) \\
& =A\left\{\left(x+\frac{B}{2 A}\right)^2-\frac{B^2}{4 A^2}+\frac{C}{A}\right\} \\
& =A\left\{\left(x+\frac{B}{2 A}\right)^2-\frac{\left(B^2-4 A C\right)}{4 A^2}\right\}
\end{aligned}
$

The above equation can be written as $
\left(x+\frac{B}{2 A}\right)^2=\frac{1}{A}\left(y+\frac{\left(B^2-4 A C\right)}{4 A^2}\right)
$

On comparing this equation with $(\mathrm{x}-\mathrm{h})^2=4 \mathrm{a}(\mathrm{y}-\mathrm{k})$ it represent a parabola with the vertex $(h, k)=\left(-\frac{B}{2 A},-\frac{B^2-4 A C}{4 A}\right)$ and axis parallel to Y - axis and latusrectum $=\frac{1}{|\mathrm{~A}|}$

Recommended Video Based on the Equation of a Parabola

Solved Examples Based on the Equation of a parabola

Example 1: The locus of the mid-point of the line segment joining the focus of the parabola $y^2=4 a x$ to a moving point of the parabola, is another parabola whose directrix is :
[JEE MAINS 2023]

Solution

$ \begin{aligned}
& \mathrm{h}=\frac{\mathrm{at}^2+\mathrm{a}}{2}, \mathrm{k}=\frac{2 \mathrm{at}+0}{2} \\
\Rightarrow & \mathrm{t}^2=\frac{2 \mathrm{~h}-\mathrm{a}}{\mathrm{a}} \text { and } \mathrm{t}=\frac{\mathrm{k}}{\mathrm{a}} \\
\Rightarrow & \frac{\mathrm{k}^2}{\mathrm{a}^2}=\frac{2 \mathrm{~h}-\mathrm{a}}{\mathrm{a}} \\
\Rightarrow & \text { Locus of }(\mathrm{h}, \mathrm{k}) \text { is } \mathrm{y}^2=\mathrm{a}(2 \mathrm{x}-\mathrm{a}) \\
\Rightarrow & y^2=2 a\left(x-\frac{a}{2}\right)
\end{aligned}
$

Its directrix is $x-\frac{a}{2}=-\frac{a}{2} \Rightarrow x=0$
Hence, the required answer is $\mathrm{x}=0$.

Example 2: If the $x$-intercept of a focal chord of the parabola $y^2=8 x+4 y+4$ is 3 ., then the length of this chord is equal to [JEE MAINS 2021]

Solution

$ \begin{aligned}
& y^2=8 x+4 y+4 \\
& (y-2)^2=8(x+1) \\
& y^2=4 a x \\
& a=2, X=x+1, Y=y-2 \\
& \text { focus }(1,2) \\
& y-2=m(x-1)
\end{aligned}
$

Put $(3,0)$ in the above line

$
m=-1
$

Length of focal chord $=16$
Hence, the answer is (16).

Example 3: Find the equation of a parabola whose latus rectum is 5 units, the axis is the line $6 x+8 y-4=0$, and the tangent at the vertex is the line

Solution: The equation of a parabola is

$
\begin{gathered}
(\mathrm{PM})^2=(\text { Latusrectum }) \cdot(\mathrm{PN}) \\
\Rightarrow\left(\frac{\mathrm{lx}+\mathrm{my}+\mathrm{n}}{\sqrt{\mathrm{l}^2+\mathrm{m}^2}}\right)^2=(\text { Latusrectum }) \cdot\left(\frac{\mathrm{mx}-\mathrm{ly}+\mathrm{r}}{\sqrt{\mathrm{m}^2+\mathrm{n}^2}}\right)
\end{gathered}
$

Using the above result, the equation of the required parabola is

$
\begin{aligned}
& \left(\frac{|6 h+8 k-4|}{\sqrt{6^2+8^2}}\right)^2=5 \cdot \frac{8 h-6 k+7}{\sqrt{6^2+8^2}} \\
& (6 h+8 k-4)^2=50(8 h-6 k+7)
\end{aligned}
$

Hence, the required answer is $(6 x+8 y-4)^2=50(8 x-6 y+7)$

Example 4: Find the value of $\lambda$ if the equation $(x+1)^2+(y-1)^2=\lambda(2 x+y+6)^2$ represents a parabola. Also, find its focus.

Solution: The equation of parabola is,

$
\begin{aligned}
& (x+1)^2+(y-1)^2=\lambda(2 x+y+6)^2=5 \lambda\left(\frac{2 x+y+6}{\sqrt{5}}\right)^2 \\
& \therefore \quad \sqrt{(x+1)^2+(y-1)^2}=\sqrt{5 \lambda} \cdot \frac{|2 x+y+6|}{\sqrt{5}}
\end{aligned}
$

This represents parabola if $\sqrt{5 \lambda}=1$

$
\Rightarrow \quad \lambda=\frac{1}{5}
$

Its focus is $(-1,1)$
Hence, the required answer is $(-1,1)$

Example 5: Length of the latus rectum of a parabola whose focus is $(2,0)$ and directrix $3 x+4 y+4=0$ is:

Solution
Length of the latus rectum $=4 \mathrm{a}$
$=2($ distance from the focus of the directrix $)$

$
=2 \times \frac{|3 \times 2+4 \times 0+4|}{\sqrt{(9+16)}}=\frac{2 \times 10}{5}=4 \mathrm{unit}
$

Hence, the answer is 4.

Summary

The parabolic curve helps us understand the relation between geometric properties and algebraic equations. It also elaborates on the practical applications of parabolas in fields such as optics, engineering, and physics. Knowledge of the Parabolic curve helps us in solving real-life problems.