Normal at t1 meets the parabola again at t2

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).

$\begin{equation}
\text { Hence it is a conic section with eccentricity e }=1 \text {. }
\end{equation}$

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

Standard equation of a parabola

If the directrix is parallel to the y-axis in the standard equation of a parabola is given as
$
y^2=4 a x
$
If the directrix is parallel to the $x$-axis, the standard equation of a parabola is given as

$
x^2=4 a y
$

Normal at t₁ meets the parabola again at t₂

A Parabola is a U- shaped plane curve where any point is at an equal distance from a fixed point and from a fixed straight line. The line perpendicular to the tangent to the curve at the point of contact is normal to the parabola. In real life, we use a parabolic antenna or parabolic microphone.

Equation of Normal at $P \equiv\left(a t_1^2, 2 a t_1\right)$ to the parabola $y^2=4 a x$ is

$
y=-t_{1 x}+2 a t_1+a t_1^3
$
It meets the parabola again at $\mathrm{Q} \equiv\left(\mathrm{at}_2^2, 2 \mathrm{at}_2\right)$

$
\begin{aligned}
& \therefore 2 \mathrm{at}_2=-\mathrm{at}_1 \mathrm{t}_2^2+2 \mathrm{at}_1+\mathrm{at}_1^3 \\
& \Rightarrow 2 a\left(t_2-t_1\right)+a t_1\left(t_2^2-t_1^2\right)=0 \\
& \Rightarrow \mathrm{a}\left(\mathrm{t}_2-\mathrm{t}_1\right)\left[2+\mathrm{t}_1\left(\mathrm{t}_2+\mathrm{t}_1\right)\right]=0 \\
& \therefore 2+\mathrm{t}_1\left(\mathrm{t}_1+\mathrm{t}_2\right)=0
\end{aligned}
$
$
t_2=-t_1-\frac{2}{t_1}
$

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Solved Questions Based on Normal at t₁ Meets the Parabola Again at t₂

Example 1: Two normals at t₁ and t₂ meet again the parabola y² = 4ax then the relation between t₁ and t₂ is

$\begin{aligned} & \text { 1) } t_1 t_2=1 \\ & \text { 2) } t_1-t_2=2 \\ & \text { 3) } t_1 t_2=2 \\ & \text { 4) } t_1+t_2=2\end{aligned}$

Solution

Normal at t1 meets the parabola again at t2 -

Equation of Normal at $\mathrm{P} \equiv\left(\mathrm{at}_1^2, 2 a t_1\right)$ to the parabola $\mathrm{y}^2=4 \mathrm{ax}$ is

$
=-\mathrm{t}_{1 \mathrm{X}}+2 \mathrm{at}_1+\mathrm{at}_1^3
$
It meets the parabola again at $\mathrm{Q} \equiv\left(\mathrm{at}_2^2, 2 \mathrm{at} 2\right)$

$
\begin{aligned}
& 2 \mathrm{at}_2=-\mathrm{at}_1 \mathrm{t}_2^2+2 \mathrm{at}_1+\mathrm{at}_1^3 \\
& \Rightarrow 2 \mathrm{a}\left(\mathrm{t}_2-\mathrm{t}_1\right)\left[2+\mathrm{at}_1\left(\mathrm{t}_2^2-\mathrm{t}_1^2\right)=0\right. \\
& \Rightarrow \mathrm{a}\left(\mathrm{t}_2-\mathrm{t}_1\right)\left[2+\mathrm{t}_1\left(\mathrm{t}_2+\mathrm{t}_1\right)\right]=0 \\
& a\left(t_2-t_1\right)=0 \\
& 2+\mathrm{t}_1\left(\mathrm{t}_1+\mathrm{t}_2\right)=0 \\
& 2=-\mathrm{t}_1-\frac{2}{\mathrm{t}_1}
\end{aligned}
$

Since the normal at $t_1$ meets the parabola at $t$, so

$
t=-t_1-\frac{2}{t_1}
$
Similarly,
Thus, $-t_1-\frac{2}{t_1}=-t_2-\frac{2}{t_2}$

$
\begin{aligned}
& \Rightarrow \quad\left(t_1-t_2\right)=\left(\frac{2}{t_1}-\frac{2}{t_2}\right)^{t_2}=\frac{2\left(t_1-t_2\right)}{t_1 t_2} \\
& \Rightarrow \quad t_1 t_2=2
\end{aligned}
$

Example 2: Let the tangent to the parabola $\mathrm{S}: \mathrm{y}^2=2 \mathrm{x}$ at the point $\mathrm{P}(2,2)$ meet the x -axis at $Q$ and normal at it meet the parabola $S$ at the point $R$. Then the area (in sq. units) of the triangle $P Q R$ is equal to :
1) $\frac{25}{2}$
2) $\frac{35}{2}$
3) $\frac{15}{2}$
4)25

Solution

$
y^2=2 x \Rightarrow a=\frac{1}{2}
$
Equation of tangent at $P$ :

$
\begin{aligned}
& T=0 \\
& y y_1=2\left(\frac{x+x_1}{2}\right) \\
& 2 y=x+2
\end{aligned}
$
For point Q , put $y=0 \Rightarrow x=-2$

$
\therefore Q(-2,0)
$

Now point $R$
If P is $\left(\mathrm{at}^2, 2 \mathrm{at}\right)$, then point R will be

$
\begin{aligned}
& \quad\left(a\left(-t-\frac{2}{t}\right)^2, 2 a\left(-t-\frac{2}{t}\right)\right) \\
& P(2,2)=\left(a t^2, 2 a t\right) \\
& \Rightarrow \quad 2=2 a t \Rightarrow 2=2\left(\frac{1}{2}\right) t \Rightarrow t=2 \\
& \therefore \quad R=\left(\frac{1}{2}\left(-2-\frac{2}{2}\right)^2, 2\left(\frac{1}{2}\right)\left(-2-\frac{2}{2}\right)\right) \\
& \quad R\left(\frac{9}{2},-3\right) \\
& \quad \Delta P Q R=\left|\frac{1}{2}\right| \begin{array}{ccc}
2 & 2 & 1 \\
-2 & 0 & 1 \\
\frac{9}{2} & -3 & 1
\end{array}||=\frac{25}{2}
\end{aligned}
$

Hence, the correct answer is option (1).

Example 3: $P$ and $Q$ are two distinct points on the parabola, $y^2=4 x$, with parameters $t$ and $t_1$ respectively. If the normal at $P$ passes through $Q$, then the minimum value of $t_1^2$ is:

1) 2

2) 4

3) 6

4) 8

Solution

Parametric coordinates of parabola -
$
\begin{aligned}
& x=a t^2 \\
& y=2 a t
\end{aligned}
$

- wherein

For the parabola.

$
y^2=4 a x
$
Parametric coordinates are $\left(a t^2, 2 a t\right)$ $t_1=-t-\frac{2}{t}($ Condition for normal chord $)$

$
t_{\text {so, }}^2=t^2+\frac{4}{t^2}+4
$
Using $A M>G M$

$
\frac{t^2+\frac{4}{t^2}}{2} \geq \sqrt{t^2 \cdot \frac{4}{t^2}}
$

Hence, the answer is the option (4)

Example 4: The normal at the point $\left(b t_1{ }^2, 2 b t_1\right)$ on a parabola meets the parabola again in the point $\left(b t_2{ }^2, 2 b t_2\right)$, then

$
\begin{aligned}
t_2 & =-t_1+\frac{2}{t_1} \\
t_2 & =t_1-\frac{2}{t_1} \\
t_2 & =t_1+\frac{2}{t_1} \\
t_2 & =-t_1-\frac{2}{t_1}
\end{aligned}
$

Solution
As we learnt in
Parametric coordinates of parabola -

$
\begin{aligned}
& x=a t^2 \\
& y=2 a t
\end{aligned}
$

- wherein

For the parabola.

$
y^2=4 a x
$

If $t_1$ and $t_2$ are the parameters

$
t_2=-t_1-\frac{2}{t_1}
$
Example 5: Circles are drawn taking any two focal chords of the parabola $y^2=4 a x$ as diameters. Then, the equation of a common chord is:
1) $a\left(t_3^2+t_4^2-t_1^2-t_2^2\right) x+a\left(t_3+t_4-t_1-t_2\right) y=0$
2) $a\left(t_3^2+t_4^2+t_1^2-t_2^2\right) x-2 a\left(t_3+t_4-t_1+t_2\right) y=0$
3) $a\left(t_3^2+t_4^2-t_1^2-t_2^2\right) x+2 a\left(t_3+t_4-t_1-t_2\right) y=0$
4) None of the above

Solution
Let $P Q$ and $R S$ be the two focal chords of the parabola and let $P, Q, R$, and $S$ be the points $t_1, t_2, t_3, t_4$ respectively, then $t_1 t_2=-1$ and $t_3 t_4=-1$. Clearly, a circle with PQ as the diameter is: $\square$

$
\begin{aligned}
\left(x-a t_1^2\right)\left(x-a t_2^2\right)+\left(y-2 a t_1\right)\left(y-2 a t_2\right) & =0 \\
\Rightarrow x^2+y^2-a\left(t_1^2+t_2^2\right) x-2 a\left(t_1+t_2\right) y-3 a^2 & =0
\end{aligned}
$
Circle with RS as the diameter is:

$
x^2+y^2-a\left(t_3^2+t_4^2\right) x-2 a\left(t_3+t_4\right) y-3 a^2=0
$

$\therefore$ Subtracting both the equations, we get the equation of the common chord as:

$
a\left(t_3^2+t_4^2-t_1^2-t_2^2\right) x+2 a\left(t_3+t_4-t_1-t_2\right) y=0
$
Hence, the answer is the option (3).