Points of intersections of the line $y=mx+c$ with the parabola $y^2=4ax$:

Three Possibilities:

1. Two points of intersection.
2. One points of intersection.
3. No points of intersection.

To find the points of intersection $y=mx+c$ and $y^2=4ax$, we solve

$(mx+c)^2=4ax$

$\Leftrightarrow m^2x^2 + 2mcx + c^2 = 4ax$

$\Leftrightarrow m^2x^2 + 2(mc-2a)x + c^2 = 0$

This is a quadratic equation in x, so it has either two real & distinct roots or equal roots or two non real complex roots.

Disriminant = $[2(mc-2a)]^2 - 4m^2c^2$

= $4 [m^2c^2 - 4amc + 4a^2 - m^2c^2]$

= $16(a^2-amc)$

$\therefore$ there are 2 points of intersection if $a^2-amc>0$

that is $a>mc$

There is one point of intersection if $a=mc$

& No points if intersection if $a

When $a=mc$ the line $y=mx+c$ intersects the parabola is only one point, and hence it is tangent to the parabola $y^2=4ax$ at the point (x,y) given by

$x= \frac{-2(mc-2a)}{2m^2} = \frac{2(mc-2mc)}{2m^2}$

$= \frac{mc}{m^2} = \frac{c}{m}$

Putting $x=\frac{c}{m}$ in the equation. $y=mx+c$ gives $y=m(\frac{c}{m}) + 2c$

So, the line $y=mx+c$ is tangent to the parabola $y^2=4ax$ at the point ($\frac{c}{m},2c$) provided $mc=a$

Length of chord of parabola $y^2=4ax$:

Length of the chord joining ($x_1,y_2$) & ($x_2,y_2$) on the parabola $y^2=4ax$ is

l = $\sqrt{(x_1 - x_2)^2 + (y_1 -y_2)^2}$

Equation of the line joining ($x_1,x_2$) & ($x_2, y_2$) is

$y-y_1 = \frac{y_1-y_2}{x_1-x_2} (x-x_1)$

$y=mx+c$, where $m = \frac{y_1-y_2}{x_1-x_2}$

$c=y_1 - (\frac{y_1-y_2}{x_1-x_2}) x_1$

Provided $x_1 \neq x_2$

Note that if $x_1=x_2$ then

$l=|y_1-y_2|$

$y_1 ^2 = 4ax , y_2 ^2 = 4ax$

$y_1=-y_2$ & $|y_1-y_2|=2\sqrt{4ax_1} = 4\sqrt{ax_1}$

$x_1$ and $x_2$ satisfy the quadratic equation.

$m^2x^2 + 2(mc-2a)x + c^2= 0$

$\therefore x_1 + x_2 = \frac{-2(mc-2a)} { m^2}$

& $x_1 x_2 = \frac{c^2}{m^2}$

$\therefore (x_1-x_2)^2 = (x_1 + x_2)^2 - 4x_1x_2$

$= \frac{4}{m^4} (mc-2a)^2 - \frac{4c^2}{m^2}$

$= \frac{4}{m^4} [m^2 c^2 - 4amc + 4a^2 - m^2c^2]$

$= \frac {16 a (a-mc)}{m^4}$

Also, $y_1-y_2 = (mx_1 + c) - (mx_2 + c) = m(x_1 - x_2)$

$\therefore$ Length of the chord, l = $\sqrt{(x_1-x_2)^2 + (y_1 - y_2)^2}$

= $\sqrt{(x_1 -x_2)^2 + m^2 (x_1 - x_2)^2}$

= $\sqrt{1+m^2} | x_1-x_2 |$

= $\sqrt{1+m^2} \frac{4}{m^2} \sqrt{a (a-mc)}$

To get the formula for l in terms of ($x_1, x_2$) & ($y_1,y_2$) we can put

$m = \frac{y_1-y_2}{x_1-x_2}$ & $c= y_1 - (\frac{y_1-y_2}{x_1-x_2}) x_1$

Equation of the tangent line to the parabola $y^2=4ax$ at a point ($x_1,x_2$):

Suppose the line has slope m then the equation of the tangent line is given by $y-y_1=m(x-x_1)$

i.e $y=mx+(y_1-mx_1)$

We know that this line is tangent to $y^2=4ax$

if $a=mc = m(y_1-mx_1)$

$\Leftrightarrow m^2 x_1 - my_1 + a =0$

$\Leftrightarrow m = \frac{y_1 \pm \sqrt{y_1 ^2 - 4ax_1}}{2x_1}$, But $y_1^2 = 4ax_1$

So, $m = \frac{y_1}{2x_1}$ This can also be written as

$m = \frac{y_1 ^2}{ax_1y_1} = \frac{4ax_1}{ax_1y_1} = \frac{2a}{y_1}$

$m = \frac{2a}{y_1}$

Equation of the tangent line is

$y=mx+(y_1-mx_1)$

= $\frac{2a}{y_1} x +y_1 - \frac{y_1}{2}$

= $\frac{2ax}{y_1} + \frac{y_1}{2}$

$\Leftrightarrow yy_1 = 2ax + \frac{y^2}{2} = 2ax + \frac{4ax_1}{2}$

$\Leftrightarrow yy_1 = 2a(x+x_1)$

Equation of the tangent line to the parabola $y^2=4ax$ at the point ($x_1,y_1$) on the parabola.

$y^2=4ax$: Equation of the parabola ($x_1,y_1$) lies on the parabola $\Rightarrow y_1^2 = 4ax_1$. we know that the slope of the tangent line at a point ($x_1,y_1$) on any curve $y=f(x)$ is given by $m=\frac{dy}{dx} |_{e(x_1,y_1)}$

$y^2=4ax \Rightarrow 2y \frac{dy}{dx} = 4a$

$\Rightarrow \frac{dy}{dx} = \frac{2a}{y} \Rightarrow m=\frac{2a}{y_1}$

$\therefore$ equation of the tangent line is

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

$\Leftrightarrow yy_1 - y_1 ^2 = 2ax-2ax_1$

But $y_1 ^2 = 4ax_1$

$\therefore yy_1 - 4ax_1 = 2ax - 2ax_1$

$\Leftrightarrow yy_1 = 2ax + 2ax_1$

$\Leftrightarrow yy_1 = 2a(x+x_1)$

Remark: The above derivation assumed

$(x_1,y_1)\neq 0$

However, if the point $(x_1,y_1)=(0,0)$ Then it is clear that the tangent line at (0,0) is the y-axis, where equation is x=0.

If we put $(x_1,y_1)=(0,0)$ in the equation obtained $yy_1 = 2a (x+x_1)$

we get 0 = 2a(x+0) i.e x=0

Hence, the equation $yy_1 = 2a (x+x_1)$ is valid for even the point $(0,0)$

general equation of tangent at $(x_1,y_1)$

Equation of the normal at point $(x_1,y_1)$ on the parabola $y^2=4ax$:

Equation of the tangent is $yy_1=2a(x+x_1)$

If $y_1 \neq 0$ then $y=\frac{2a}{y_1} (x+x_1)$

$\therefore$ Slope of tangent = $\frac{2a}{y_1}$

$\Rightarrow$ Slope of normal, m = $\frac{-y_1}{2a}$

$\therefore$ Equation of normal is

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

In terms of the slope m:

$m=\frac{-y_1}{2a}$ i.e $y_1 = -2am$

$\therefore x_1 = \frac{y_1 ^2} {4a} = \frac{4a^2 m^2 }{4a} = am^2$

Putting in the equation. $y-y_1 = \frac{-y_1}{2a} (x-x_1)$,

we get

$y+2am=m(x-am^2)$

$\Leftrightarrow y=mx-2am-am^3$

equation of the normal at the points $(x_1,y_1) = (am^2,-2am)$

AT: subtangent

AN: subnormal

Equation of line PT is

$yy_1 = 2a(x+x_1)$

Putting $y=0$ gives $x=-x_1$

$\therefore T \equiv (-x_1,0)$

Subtangent $AT = 2x_1$

$\triangle TAP$ is similar to $\triangle PAN$

$\Rightarrow \frac{AN}{AP} = \frac{AP}{AT} \Rightarrow AN = \frac{AP^2}{AT} = \frac{y_1^2}{2x_1} = \frac{4ax_1}{2x_1} = 2a$

$\therefore$ Subnormal, $AN = 2a$

Which is a constant

Thank you