Properties 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 $x y=4 \mathrm{a}^2$