The equation of tangent to the parabola `y^2 = 4ax` at `(x_1, y_1) ` is

`yy_1=2a(x+x_1)` ...................(1)

Since `m` is the slope of the tangent then

`m=(2a)/(y_1)` or `y_1=(2a)/m`

Since `(x _1, y_1)` lies on `y^2 = 4ax` therefore

`y_1^2=4ax_1` or `(4a^2)/(m^2)=4ax_1` `=> x_1=a/(m^2)`

Substituting the values of `x_1` and `y_1` in ( 1), we get

`y=mx+a/m`......................(2)

Thus, `y=mx+a/m` is a tangent to the parabola `y^2 =4ax` for all values of `m`, where `m` is the slope of the tangent and the co-ordinates of the point of contact is

`(a/(m^2),(2a)/m)`.

Thus `y =mx + c` is the tangent of `y^2 = 4ax` for all values of `m` if only if `c = a/m` and `(y-k)=m(x-h)+a/m` is tangent to the parabola `(y - k)^2 = 4a (x - h).`

Let the parabola be `y^2= 4ax`............................(1)

Let `P(alpha, beta)` be the given point

The equation of a tangent to parabola `(1)` is `y = mx +a/m`....................(2)

lf line `(2)` passes through `P (alpha, beta)` , then `beta = m alpha + a/m` or `m^2 alpha - beta m + a = 0` .............. (3)

There will be two tangents to parabola `(1)` from `P (alpha , beta)` if roots of equation `(3)` are real and distinct

i.e., `D > 0` i.e. if `beta^2 - 4 alpha a > 0 => P(alpha , beta)` lies outside parabola `(1)`.

We can also find the angle between two tangents from point `P( alpha, beta)` using the formula

`tan theta=|(m_1-m_2)/(1+m_1m_2)|`

We have to find the equation of tangent to the parabola `y^2 = 4ax` at the point `(at^2, 2at)` or `'t'`

Since the equation of tangent of the parabola `y^2 = 4ax` at` (x_1, y_1)` is

`yy_1 = 2a(x + x _1 )` ................. (1)

replacing `x_1` by `at^2` and `y_1` by `2at`, then `(1)` becomes

`y(2at) = 2a(x + at^2)`

`ty =x + at^2`

Let the given parabola be `y^2 = 4ax` and two points on the parbaola are

`P = (at_1^2, 2at_1)` and `Q = (at_2^2, 2at_2)`

Equation of tangents at

`P (at_1^2, 2at_1)` and `Q(at_2^2, 2at_2)`

are `t_1y = x + at_1^2` ..........................(1)

and `t_2y = x + at_2^2` ..................... (2)

Solving these equations we get

`x = at_1t_2 , y = a(t_1 + t_2)`

Thus, the co-ordinates of the point of intersection of tangents at

`P(at_1^2, 2at_1)` and `Q(at_2^2 , 2at_2)` are `R(at_1t_2 , a(t_1 + t_2))`.

e.g. Find the equation of the chord of the parabola y2 = 12x which is bisected at the point (5, –7).

Solution:

Here `(x_1, y_1) = (5, –7),` and `y_2 = 12x = 4ax ⇒ a = 3.`

The equation of the chord is `S_1 = T`

or `(y_1)^2 – 4ax_1 = yy_1 – 2a(x + x_1)` or `(–7)^2 – 12.5 = y(–7) – 6 (x + 5)`.

Or` 6x + 7y + 19 = 0.`

Since the equation of the tangent to the parabola `y^2 = 4ax` at `(x_1, y_1)` is

`yy_1= 2a(x+x_1)` .......... (1)

The slope of the tangent at ` (x_1, y_1) = (2a)/(y_1)`

`:.` Slope of the normal at `(x_1, y_1) = - (y_1)/(2a)`

Hence the equation of normal at `(x_1, y_1)` is

`y - y_1 = -(y_1)/(2a)(x - x_1)`

The equation of normal to the parabola `y^2 = 4ax` at `(x_1 ,y_1)` is

`y-y_1=-(y_1)/(2a)(x-x_1)`...............(1)

Since `m` is the slope of the normal

then `m=-(y_1)/(2a)` or `y_1=-2 am`

Since `(x_1 ,y_1)` lies on `y^2 = 4ax` therefore

`y_1^2=4ax_1` or `4a^2m^2=4ax_1`

`:. x_1 = am^2`

Substituting the values of `x_1` and `y_1` in (1) we get

`y + 2am = m(x - am^2)`...........................(2)

Thus, `y= mx - 2am - am^3` is a normal to the parabola `y^2 = 4ax` where `m` is the slope of the normal. The co-ordinates of the point of contact are `(am^2, - 2am)`.

Hence `y = mx+c` wil lbe normal to parabola. lf and only if `c = - 2am - am^3`

Equation of normal of the parabola `y^2 = 4ax` at `(x_1,y_1)` is

`y-y_1=(y_1)/(2a)(x-x_1)` ........................(1)

Replacing `x_1` by `at^2` and `y_1` by `2at` then (l) becomes

`y - 2at = - t(x - at)^2`

or `y = - tx + 2at + at^3`

(a) Point of intersection of normals at any two points on the parabola :

Let the points `P(at_1^2, 2at_1)` and `Q (at_2^2, 2at_1)` lie on the parabola `y^2 = 4ax`

The equations of the nonnals at `P(at_1^2, 2at_1)` and `Q(at_2^2, 2at_1)` are

`y =-t_1x + 2at_1+ at_1^3` ....................(1)

and `y =- t_2x + 2at_2 + at_2^3` .................(2)

Hence point of intersection of above normals will be obtained by solving (1) and (2), we get

`x = 2a + a(t_1^2 + t_2^2 + t_1t_2)`

`y =- at_1 t_2(t_1 + t_2)`

lf `R` is the point of intersection then it is

`R = [2a + a(t_1^2 + t_2^2 + t_1t_2) , - at_1t_2 (t_1 + t_2)]`

(b) Relation between `t_1 ` and `t_2` if normal at `t_1` meets the parabola again at `t_2` :

Let the parabola be `y^2 = 4ax`, equation of normal at `P(at_1^2, 2at_1)` is

`y =- t_1x+2at_1 + at_1^3` .............................(1)

Since normal meet the parabola again at `Q(at_2^2, 2at_2)`

`:. 2at_2 =- at_1 t_2^2 + 2at_1 + at_2^3`

`=> 2a(t_2 - t_1) + at_1 (t_2^2 - t_1^2) = 0`

`=> a(t_2 - t_1) [2 + t_1(t_2 + t_1)] = 0`

`a(t_2 - t_1) ne 0` ( `∵ t_1 ` and `t_2` are different)

`:. 2 + t_1(t_2+ t_1) = 0`

`:. t_2 =-t_1-2/(t_1)`


(c) If normal to the parabola `y^2 = 4ax` drawn at any point `(at^2, 2at)` meet the parabola at `t_3` then

`t_3=-t-2/t`

`=> t^2+t t_3 +2 =0` .................(1)

It has two roots `t_1 ` & `t_2`. Hence there are two such point `P(t_1)` & ` Q(t_2)` on the parabola from where normals
are drawn and which meet parabola at `R(t_3)`

`=> t_1 + t_2 = - t_3` & `t_1t_2 = 2`

Thus the line joining `P(t_1)` & `Q(t_2) ` meet `x`-axis at `(- 2a, 0)`

Maximum three normals can be drawn from a point to a parabola and their feet (points where the normal meet the parabola) are called co-normal points.

Let `P(h, k)` be any given point and `y^2 = 4ax` be a parabola.

The equation of any normal to `y^2 = 4ax` is

`y = mx - 2am - am^3`

If it passes through `(h, k)` then

`k = mh - 2am - am^3`

`=> am^3 + m(2a - h) + k = 0` ......... (i)

This is a cubic equation in m, so it has three roots, say

`m_1,m_2` and `m_3` .

`:. m_1 +m_2 +m_3 = 0` .......... (ii)

`m_1+m_2+m_3=0` .......................(iii)

`m_1m_2m_3=-k/a` ...........................(iv)

Hence for any given point `P(h, k)` , (i) has three real or imaginary roots. Correspoinding to each of these three roots, we have each normal passing through `P(h, k)`. Hence we have three normals `PA, PB` and `PC` drawn through `P` to the parabola.

Points `A, B, C` in which the three normals from `P(h, k)` meet the parabola are called co-normal points.

Properties of co-normal points :

(I) The algebraic sum of the slopes of three concurrent normals is zero. This follows from equation (ii).

(2) The algebraic sum of ordinates of the feets of three normals drawn to a parabola from a given point is
zero.

Let the ordiantes of `A, B, C` be `y_1, y_2, y_3` respectively then

`y_1 = -2` am, `y_2 = -2am_2` and `y_3 = -2am_3`

.-. Algebraic sum of these ordinates is

`y_1 + y_2 + y_3 = - 2am_1- 2am_2 - 2am_3`

`= - 2a(m 1 + m2 + m3)`

`= - 2a xx 0` (from equation (ii)}

`= 0`

(3) If three normals drawn to any parabola `y^2 = 4ax` from a given point `(h, k)` is real then `h > 2a`.
When normals are real, then all the three roots of equation

(i) are real and in that case

`m_1^2+ m_2^2+ m_3^2 > 0` (for any values of `m_1,m_2,m_3` )

`=> (m_1 + m_2 + m_3)^2 - 2 (m_1m_2 +m_2m_3 +m_3m_1) > 0`

`=> (0)^2 - (2(2a - h))/a > 0`

`=> h - 2a > 0`

Or `h > 2a`


(4) The centroid ofthe triangle formed by the feet ofthe three normals lies on the axis of the parabola.

lf `A(x_1 ,y_1) , B(x_2, y_2)` and `C(x_3, y_3)` be vertices of `triangleABC` , then its centroid is

`((x_1+x_2+x_3)/3 , (y_1+y_2+y_3)/3) = ((x_1+x_2+x_3)/3 , 0)`

Since `y_1 + y_2 + y_3 = 0` (from result-2). Hence the centroid lies on the x-axis, which is the axis of the
parabola also

`(x_1+x_2+x_3)/3 =1/3 (am_1^2+ am_2^2 +am_3^2) a/3 (m_1^2 +m)_2^2 + m_3^2)`

`=a/3{(m_1+m_2 +m_3)^2 - 2(m_1m_2+m_2m_3+m_3m_1)}`

`=a/3 {(0)^2- 2((2a-h)/a)} = (2h-4a)/3`

Centroid of `triangleABC` `((2h-4a)/3 , 0)`