Properties 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 e = 1. In real life, we use Parabolas in bridges, telescopes, satellites, etc.

In this article, we will cover the concept of Some Standard Property 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 twelve questions have been asked on JEE MAINS( 2013 to 2023) from this topic including in two 2021.

This Story also Contains

1. What is Parabola?
2. Standard equation of a parabola
3. Derivation of Standard equation of a parabola
4. Solved Examples Based on Some Standard Property of 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).

Hence it is a conic section with eccentricity e = 1.

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

Standard equation of a parabola

The required equation of a standard parabola is

$
y^2=4 a x
$

Derivation of Standard equation of a parabola

Let focus of parabola is $S(a, 0)$ and directrix be $x+a=0$
$P(x, y)$ is any point on the parabola.
Now, from the definition of the parabola,

$
\begin{array}{cc}
& \mathrm{SP}=\mathrm{PM} \\
\Rightarrow & \mathrm{SP}^2=\mathrm{PM}^2 \\
\Rightarrow & (\mathrm{x}-\mathrm{a})^2+(\mathrm{y}-0)^2=(\mathrm{x}+\mathrm{a})^2 \\
\Rightarrow & \mathrm{y}^2=4 \mathrm{ax}
\end{array}
$

Some Standard Properties of Parabola

1. The portion of a tangent to a parabola intercepted between the directrix and the curve subtends a right angle at the focus.

The equation of the tangent to the parabola $y^2=4 a x$ at $P\left(a t^2, 2 a t\right)$ is

$
\text { ty }=x+a t^2
$

Let Eq. (i) meet the directrix $x+a=0$ at $Q$
then coordinates of $Q$ are $\left(-a, \frac{a t^2-a}{t}\right)$, also focus $S$ is $(a, 0)$

.
Slope of $S P=\frac{2 a t-0}{a t^2-a}$

$
=\frac{2 \mathrm{t}}{\mathrm{t}^2-1}=\mathrm{m}_1 \quad[\text { say }]
$

and $\quad$ slope of $S Q=\frac{\frac{a t^2-a}{t}-0}{-a-a}=\frac{t^2-1}{-2 t}=m_2$

$
\therefore \quad m_1 m_2=-1
$

i.e. $S P$ is perpendicular to $S Q$ i.e. $\angle P S Q=90^{\circ}$
2. The tangent at a point $P$ on the parabola $y^2=4 a x$ is the bisector of the angle between the focal radius $S P$ and the perpendicular from $P$ on the directrix.

Let $\mathrm{P} \equiv\left(a t^2, 2 a t\right), \mathrm{S} \equiv(a, 0)$
Equation of SP is :

$
\begin{array}{rr}
& y-0=\frac{2 a t-0}{a t^2-a}(x-a) \\
\Rightarrow & 2 t x+\left(1-t^2\right) y+(-2 a t)=0
\end{array}
$
Equation of PM is :

$
y-2 a t=0 \ldots
$
Angle bisectors of (i) and (ii) are:

$
\begin{aligned}
& \frac{y-2 a t}{\sqrt{0+1}}= \pm \frac{2 t x+\left(1-t^2\right) y-2 a t}{\sqrt{4 t^2+\left(1-t^2\right)^2}} \\
& \Rightarrow y-2 a t= \pm \frac{2 t x+\left(1-t^2\right) y-2 a t}{1+t^2} \\
& \Rightarrow t y=x+a t^2 \text { and } t x+y=2 a t+a t^3 \\
& \Rightarrow \text { tangent and normal at P are bisectors of SP } \\
& \text { and PM. }
\end{aligned}
$
3. The foot of the perpendicular from the focus on any tangent to a parabola lies on the tangent at the vertex.

Equation of tangent at $P\left(a t^2, 2 a t\right)$ on the parabola $y^2=4 a x$ is

$
\begin{array}{ll}
& \text { ty }=\mathrm{x}+\mathrm{at}^2 \\
\Rightarrow & \mathrm{x}-\mathrm{ty}+\mathrm{at}^2=0
\end{array}
$
Now, the equation of line through $S(a, 0)$ and perpendicular to Eq. (i) is

$
\mathrm{tx}+\mathrm{y}=\lambda
$
This eq passes through (a, 0)

Hence, the point of intersection of Eq. (i) and (ii) lies on $x=0$, which is the equation of tangent at the vertex

$
\begin{aligned}
& \therefore \quad \mathrm{t}(\mathrm{a})+(0)=\lambda \\
& \therefore \quad \text { Equation } t x+y=t a \quad \text { or } \quad t^2 x+t y-a t^2=0 \quad \ldots \text { (ii) } \\
& \text { adding equation (i) and equation (ii) we get } \\
& x\left(1+t^2\right)=0 \\
& \Rightarrow \quad x=0 \quad\left[\because 1+t^2 \neq 0\right]
\end{aligned}
$

4. If S is the focus of the parabola and tangent and normal at any point P meets its axis in T and G respectively, then ST = SG = SP

Equation of tangent and Normal at $P\left(a t^2, 2 a t\right)$ on the parabola $y^2=4 a x$ are

$
t y=x+a t^2 \quad \text { and } y=-t x+2 a t+a t^2, \text { respectively. }
$
Since, tangent and normal meet its axis in $T$ and $G$.
$\therefore$ Coordinates of $T$ and $G$ are $\left(-a t^2, 0\right)$ and $\left(2 a+a t^2, 0\right)$ respectively

$
\therefore \quad \begin{aligned}
S P & =P M=a+a t^2 \\
S G & =V G-V S=2 a+a t^2-a \\
& =\mathrm{a}+\mathrm{at}^2
\end{aligned}
$

and $S T=V S+V T=a+a t^2$
Hence, $S P=S G=S T$

Solved Examples Based on Some Standard Property of Parabola

Example 1: Let g-parabola $P$ be such that its vertex and focus lie on the positive $x$-axis at a distance of 2 and 4 units from the origin, respectively. If tangents are drawn from $O(0,0)$ to the parabola $P$ which meets $P$ at $S$ and $R$ then the area (in sq. units) of $\triangle S O P$ P equal to
[JEE MAINS 2021]
Solution

$
\text { vertex }=(2,0), \text { Focus },(4,0) \Rightarrow a=2
$

So y -axis is directrix and thus Rs should pass through the focus $S^{\prime}(4,0)$

$
\begin{aligned}
& R S=4 a=8 \\
& O S^{\prime}=4
\end{aligned}
$

$
\text { area of } O R S=\frac{1}{2} \cdot 4 \cdot 8=16 \text { units }
$

Hence the answer is 16 units
Example 2: If two tangents drawn from a point $P$ to the parabola $y^2=16(x-3)$ are at right angles, then the locus of the point P is :
[JEE MAINS 2021]
Solution: The locus is directrix of Parabola

So equation of directrix is $x+1=0$
Hence, the answer is $x+1=0$
Example 3: The angle between the focal chord and the normal passing through point $P$ on the parabola $y^2=4 a x$ is $60^{\circ}$. Then the slope of the tangent at point $P$ is
Solution

|PS is the focal chord and PN is the normal
By property

$
S P=S T=S N
$

given that

$
\begin{aligned}
& \angle S P N=60^{\circ} \\
& \therefore \quad \angle S P T=30^{\circ} \\
& \text { since, } \quad S T=S P, \angle P T S=30^{\circ} \\
& \therefore \text { slope of tangent }=\tan 30^{\circ}=\frac{1}{\sqrt{3}}
\end{aligned}
$

Hence, the answer is $1 / \sqrt{3}$

Example 4: The radius of the circle that passes through the origin and touches the parabola $\mathrm{y}^2=4 \mathrm{ax}$ at the point $(\mathrm{a}, 2 \mathrm{a})$ is
Solution Equation of the tangent of the parabola at $(\mathrm{a}, 2 \mathrm{a})$ is

$
\mathrm{y} \cdot 2 \mathrm{a}=2 \mathrm{a}(\mathrm{x}+\mathrm{a})
$

i.e., $y-x-a=0$

The equation of the circle touching the parabola at $(\mathrm{a}, 2 \mathrm{a})$ is

$
(x-a)^2+(y-2 a)^2+\lambda(y-x-a)=0
$

Since, it passes through $(0,0)$, therefore

$
\begin{aligned}
& \mathrm{a}^2+4 \mathrm{a}^2+\lambda(-\mathrm{a})=0 \\
& \Rightarrow \quad \lambda=5 \mathrm{a}
\end{aligned}
$

Thus required circle is

$
x^2+y^2-7 a x+a y=0
$

It's radius $=\sqrt{\frac{49}{4} \mathrm{a}^2+\frac{\mathrm{a}^2}{4}}=\frac{5}{\sqrt{2}} \mathrm{a}$.
Hence, the answer is $\frac{5}{\sqrt{2}} \mathrm{a}$

Example 5: Parabolas $y^2=4 a\left(x-c_1\right)$ and $x^2=4 a\left(y-c_2\right)$ where $c_1$ and $c_2$ are variable, are such that they touch each other. Locus of their point of contact is
Solution: Let $\mathrm{P}(\mathrm{x}, \mathrm{y})$ be the point of contact.
At $P$ both of them must have the same slope.
We have $\underbrace{2 y \frac{\frac{d y}{d x}}{d x}}=4 a, 2 x=4 a^{\frac{d y}{d x}}$
Eliminating $\frac{d y}{d x}$, we get,

$
\mathrm{xy}=4 \mathrm{a}^2
$

Hence, the answer is $x y=4 \mathrm{a}^2$