Parametric form of the parabola $y^2=4ax$

If we put $y=t \in \R$,

then, $x = \frac{y^2}{4a} = \frac{t^2}{4a}$

So, any general point on the parabola can be written as $x=\frac{t^2}{4a}$ and $y=t$

Instead of $y=t$ if we use $y=2at$

then $x=\frac{(2at)^2}{4a} = \frac{4a^2t^2}{4a} = at^2$

We will use $x=at^2,y=2at$ as the parametric form of the parabola $y^2=4ax$.

i.e any general point of the parabola $y^2=4ax$ can be written as $(at^2, 2at)$.

As t various between $-\infty$ and $+\infty$, we get all points on the parabola.

Also, there is a unique point $(at^2, 2at)$ on the parabola for any given $t \in \R$.

Equation of tangent & normal in parametric form:

Recall : equation of tangent at $(x_1, y_1)$ is given by

$y y_1=2 a(x+x_1)$

Equation of normal at $(x_1, y_1)$ is given by

$y-y_1=\frac{-y_1}{2 a} (x-x_1)$

Using $(x_1, y_1)=(a t^2, 2 a t)$,

we get Equation of tangent:

$y.2 a t=2 a(x+a t^2)$

that is $t y=x+a t^2$

$y$-2at $=-\frac{2 a t}{3 a}(x-a t^2)$

ie. $y-2 a t=-t x+a t^3$

ie. $y+t x=2 a t+a t^3$

slope $=-t$.

Equation of the chord joining $P \equiv (at_1^2, 2at_1)$

and $Q \equiv (at_2^2, 2at_2)$

$y-2at_1 = \frac{2at_1-2at_2}{at_1^2-at_2^2} (x-at_1^2)$

$\Leftrightarrow y-2at_1 = \frac{2}{t_1+t_2} (x-at_1^2)$

$\Leftrightarrow (t_1+t_2)y - 2at_1 (t_1+t_2) = 2(x-at_1^2)$

$\Leftrightarrow y(t_1+t_2) = 2at_1^2 + 2a t_1 t_2 + 2x - 2at_1^2$

$\Leftrightarrow y(t_1+t_2) = 2(x+at_1t_2)$

Point of intersection of tangents at two different point on the parabola $y^2=4 a x$.

Let $P \equiv(a t_1^2, 2 a t_1)$ & $Q \equiv(a t_2^2, 2 a t_2)$

Tangent at $P: \quad t, y=x+a t_1^2$

Target at Q : $t_2 y=x+a t_2^2$

$(i)-(ii) \Rightarrow (t_1-t_2) y=a(t_1^2-t_2^2)$

$\Rightarrow y=a(t_1+t_2)$

$\therefore x =t_1 y-a t_1^2=a t_1(t_1+t_2)-a t_1^2 = a t_1 t_2$

Point of intersection is $(a t_1 t_2, a(t_1+t_2))$

If the tangents at point $P$ and $Q$ on a parabola meet at point $T$, prove that

(i) $T P$ and $T Q$ subtend equal angle at the focus S .

(ii) $S T^2=S P \cdot S Q$

(iii) $\triangle S P T$ is similar to $\triangle S T Q$.

let $P \equiv(a t_1^2, 2 a t_1)$ & $Q \equiv(a t_2^2, 2 a t_2)$ By the previous formula for the point of intersection of tangent lives at $P$ & $Q$, we have

$T \equiv(a t_1 t_2, a(t_1+t_2))$

Equation of the line $S P$

$y-0=\frac{2 a t_1-0}{a t_1^2-a}(x-a)$

$[\therefore S \equiv (a,0) and P \equiv (at_1^2,2at_1)]$

$\Leftrightarrow y=\frac{2 t_1^2}{t_1^2-1}(x-a)$

$\Leftrightarrow (t_1^2-1) y-2 t_1 x+2 a t_1=0$

Now let M be the foot of perpendicular from T to the Joining S & P.

Recall: The perpendicular distance of a point $(x_1,y_1)$ to the line $ax+bx+c=0$ is given by

$\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}$

$\therefore TM = \frac{|(t_1^2-1) a(t_1+t_2)-2t_1 a t_1t_2+2at_1|}{\sqrt{(t_1^2-1)^2 + (2t_1)^2}}$

$= \frac{a|(t_1^2-1)(t_1+t_2)-2t_1^2t_2+2t_1|}{\sqrt{(t_1^2+1)^2}}$

$\therefore T M =\frac{a|t_1^3-t_1+t_1^2 t_2-t_2-2 t_1^2 t_2+2 t_1|}{t_1^2+1}$

$=\frac{a|t_1^3-t_1^2 t_2+(t_1-t_2)|}{t_1^2+1}$

$=\frac{a|(t_1-t_2)(t_1^2+1)|}{(t_1^2+1)}=a|t_1-t_2|$

Since this is symmetric w.r.t. $t_1$ & $t_2$,

$T M’=T M=a|t_1-t_2|$

$TM=TM'$

$\triangle T M S$ and $\triangle T M’ S$ are congruent

$\Rightarrow \angle T S M=\angle T S M’$ ie. $\angle T S P=\angle T S Q$.

This proves (i).

$ST^2 = (SP)(SQ)$

$SP^2 = (at_1^2-a)^2 + (2at_1-0)^2 = a^2(t_1^2-1)^2+4a^2t_1^2 = a^2 (t_1^2+1)^2$

$SP = a(t_1^2+1)$

Similarly $SQ = a(t_2^2+1)$

$ST = (at_1t_2-a)^2 + (a(t_1+t_2)-0)^2$

$= a^2[(t_1t_2-1)^2+(t_1+t_2)^2]$

$= a^2 [t_1^2t_2^2+1-2t_1t_2+t_1^2+t_2^2+2t_1t_2]$

$= a^2 (t_1^2 +1)(t_2^2+1)$

$= (SP)(SQ)$, which proves (ii)

Next, to show $\Delta SPT \sim \Delta STQ$.

We have from (i) & (ii)

$\angle PST = \angle QST$

and $\frac{ST}{SP}=\frac{SQ}{ST}$

$\therefore \Delta SPT$ is similar to $\Delta STQ$.

prove that the area of the triangle formed by three points on a parabola is twice the area of the triangle formed by the tangents at these points.

Solution. We can assume that the equation of the parabola is $y^2=4 a x$ (by choosing the vertex to be at $(0,0)$ and to axis as the $x$-axis.)

let $P \equiv (a t_1^2, 2 a t_1)$

$Q \equiv (a t_2^2, 2 a t_2)$

$R \equiv (a t_3^2 ,2 a t_3)$

be any 3 points on the parabola.

Recall: Area of the triangle formed by $(x_1,y_1),(x_2,y_2)$ & $(x_3,y_3)$

Area of $\Delta PQR$ = area of trap. PABR + area of trap. RBCQ - area of trap. PACQ

$= \frac{1}{2} (y_1+y_2) (x_3-x_1) + \frac{1}{2} (y_2+y_3) (x_2-x_3) - \frac{1}{2} (y_1+y_2) (x_2-x_1)$

$\Delta = \frac{1}{2} [x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

=$\frac{1}{2}\left|\begin{array}{ccc} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\x_2 & y_2 & 1\end{array}\right|$

Then area of $\triangle P Q R$

$=\frac{1}{2} [a t_1^2(2 a t_2-2 a t_3)+a t_3^2 (2 a t_3-2 a t_1) + a t_3^2(2 a t_1-2 a t_2)]$

$=a^2[t_1^2(t_2-t_3)+t_2^2(t_3-t_1)+t_3^2(t_1-t_2$

$=-a^2(t_1-t_2)(t_2-t_3)(t_3-t_1)$ (eqn-1)

Points of intersection of tangent are

$(a t_1 t_3, a(t_1+t_2))$, $(a t_2 t_3, a(t_2+t_3))$, $(a t_3 t_1, a(t_3+t_1))$

Area of $\Delta$ formed by these tangent

$= \frac{1}{2} [at_1t_2 (at_2-at_1) + at_2t_3 (at_3-at_2) + at_3t_1 (at_1-at_3)]$

$= \frac{1}{2} a^2 (t_1-t_2)(t_2-t_3)(t_1-t_3)$

$= \frac{1}{2} \text{(area of PQR)}$

Thank you