Some Standard Property of Parabola - Practice Questions & MCQ

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

$
\mathrm{ty}=\mathrm{x}+\mathrm{at}^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 $(a, 0)$

$
\begin{aligned}
\therefore \quad \text { 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 }]
\end{aligned}
$

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 y² = 4ax is the bisector of the angle between the focal radius SP 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{aligned}
& 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{aligned}
$
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
\end{aligned}
$

$\Rightarrow \quad$ tangent and normal at P are bisectors of SP and PM.

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} 
& \mathrm{ty}=\mathrm{x}+\mathrm{at}^2 \\
\Rightarrow \quad & \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)

$
\therefore \quad \mathrm{t}(\mathrm{a})+(0)=\lambda
$

$\therefore \quad$ Equation $t x+y=t a \quad$ or $\quad t^2 x+t y-a t^2=0 \quad \ldots$ (ii) adding equation (i) and equation (ii) we get

$
\Rightarrow \quad \begin{aligned}
x\left(1+t^2\right) & =0 \\
x & =0
\end{aligned} \quad\left[\because 1+t^2 \neq 0\right]
$

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

4. If S is the focus of the parabola and tangent and normal at any point P meet 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 $\quad S T=V S+V T=a+a t^2$
Hence, $S P=S G=S T$