Normal in point and Parametric and Slope form of parabola - Practice Questions & MCQ

Normal in point form of parabola 

Point Form

The equation of the Normal at the point $P\left(x_1, y_1\right)$ to a Parabola $y^2=4 a x$ is $y-y_1=-\frac{y_1}{2 a}\left(x-x_1\right)$

Proof:

The equation of tangent to the parabola $y^2=4 a x$ at $\left(x_1, y_1\right)$ is $y_1=2 a\left(x+x_1\right)$.
Slope of the tangent at $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ is $\frac{2 \mathrm{a}}{\mathrm{y}_1}$ slope of normal at $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ become $-\frac{\mathrm{y}_1}{2 \mathrm{a}}$
$\therefore$ Equation of normal at $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ is $\mathrm{y}-\mathrm{y}_1=-\frac{\mathrm{y}_1}{2 \mathrm{a}}\left(\mathrm{x}-\mathrm{x}_1\right)$

Normal in Parametric Form of Parabola

The equation of normal to the parabola $y^2=4 a x$ at the point $\left(\mathrm{at}^2, 2 \mathrm{at}\right)$ is $\mathrm{y}+\mathrm{tx}=2 \mathrm{a} t+\mathrm{at}^3$

Proof:
The equation of the Normal at the point $\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right)$ to a Parabola $\mathrm{y}^2=4 \mathrm{ax}$ is

$
\begin{aligned}
& y-y_1=-\frac{y_1}{2 a}\left(x-x_1\right) \\
& \text { replace } x_1 \rightarrow a t^2, y_1 \rightarrow 2 a t \\
& y-2 a t=-t\left(x-a t^2\right) \Rightarrow y+t x=2 a t+a t^3
\end{aligned}
$

NOTE:
$\begin{array}{c||cc} \\\text { {Equation of Parabola} } & {\text { Coordinate }} & {\text { Tangent Equation }} \\\\\hline \hline \\y^{2}=4ax & {\left(at^{2}, 2 a t\right)} & { y+tx=2at+a t^{3}}\\ \\ y^{2}{=-4 a x} & {\left(-a t^{2}, 2 a t\right)} & { y-tx=2at+a t^{3}} \\\\ x^{2} {=4 a y} & {(2 a t, a t^2)} & { x+ty=2at+a t^{3}} \\ \\ x^{2} {=} {-4 a y} & {\left(2 a t,-at^{2}\right)} & { x-ty=2at+a t^{3}} \\ \\ \hline\end{array}$

Normal in Slope Form of Parabola

The equation of the Normal at the point $P\left(x_1, y_1\right)$ to a Parabola $y^2=4 a x$ is

$
\mathrm{y}-\mathrm{y}_1=-\frac{\mathrm{y}_1}{2 \mathrm{a}}\left(\mathrm{x}-\mathrm{x}_1\right)
$

$m$ is the slope of the tangent, then

$
\mathrm{m}=-\frac{\mathrm{y}_1}{2 \mathrm{a}} \Rightarrow \mathrm{y}_1=-2 \mathrm{am}
$

$\left(\mathrm{x}_1, \mathrm{y}_1\right)$ lies on the paarabola $\mathrm{y}^2=4 \mathrm{ax}$

$
\begin{aligned}
& \mathrm{y}_1^2=4 \mathrm{ax}_1 \Rightarrow(2 \mathrm{am})^2=4 \mathrm{ax}_1 \\
& \therefore \mathrm{x}_1=\mathrm{am}^2
\end{aligned}
$

put the value of $\mathrm{x}_1$ and $\mathrm{y}_1$ in the equation $\mathrm{y}-\mathrm{y}_1=-\frac{\mathrm{y}_1}{2 \mathrm{a}}\left(\mathrm{x}-\mathrm{x}_1\right)$ we get

$
\mathrm{y}=\mathrm{mx}-2 \mathrm{am}-\mathrm{am}^3
$

which is equation of normal of the parabola in slope form

TIP

If $\mathrm{c}=-2 \mathrm{am}-\mathrm{am}^3$, then $\mathrm{y}=\mathrm{mx}+\mathrm{c}$ is the equation of normal of the parabola $\mathrm{y}^2=$ 4 ax

TIP

When the vertex of the parabola at (h, k)

Point of Intersection of Normal of a Parabola

Let the equation of parabola be $y^2=4 a x$

Two points, $P \equiv\left(a t_1^2, 2 a t_1\right)$ and $Q \equiv\left(a t_2^2, 2 a t_2\right)$ on the parabola $y^2=4 a x$.
Then, equation of Normal; at $P$ and $Q$ are

$
\begin{aligned}
& y_1=-t_1 x+2 a t_1+a t_1^3 \\
& y_2=-t_2 x+2 a t_2+a t_2^3
\end{aligned}
$

solving (i) and (ii) we get,

$
x=2 a+a\left(t_1^2+t_2^2+t_1 t_2\right), y=-a t_1 t_2\left(t_1+t_2\right)
$
If $R$ is the point of intersection of two normal then,

$
\mathrm{R} \equiv\left[2 \mathrm{a}+\mathrm{a}\left(\mathrm{t}_1^2+\mathrm{t}_2^2+\mathrm{t}_1 \mathrm{t}_2\right),-\mathrm{at}_1 \mathrm{t}_2\left(\mathrm{t}_1+\mathrm{t}_2\right)\right]
$