In this page we are going to see how to get the equations of normal and chord of contacts to the given parabola.

Equation of tangent to the parabola at the point (x₁,y₁) is

yy₁=2a(x+x₁)

Now we will see how we will get the equation of normal from the equation of tangent.

Slope of the tangent is 2a/y₁

Slope of the perpendicular line to the tangent is -y₁/2a.

Equation of normal at (x₁,y₁) is

y-y₁= -y₁/2a(x-x₁)

2ay-2ay₁=-xy₁+x₁y₁

xy₁+2ay=2ay₁+x₁y₁.

**Example:**

** Find the equation of normal to the parabola y****²=x at (4,-2).**

**Solution:**

**Equation at ****(x₁,y₁) is **xy₁+2ay=2ay₁+x₁y₁.

Here **(x₁,y₁) is (4,-2).**

**The equation of the given parabola is ****y****²=x.**

** 4a = 1**

** a = 1/4**

**The required equation is **

** x(-2)+2(1/4)y=2(1/4)(-2)+4(-2)**

** -2x+**½y = -1-8

Multiplying the whole equation by -2

4x-y =18

** 4x-y-18=0** is the required equation**.**

__ Chord of contact__:

Equation of chord of contact of tangents from ** ****(x₁,y₁) to the parabola**** y****²=4ax.**

Let the tangents from the point P(x₁,y₁) touch the parabola at Q(x₂,y₂) and R(x₃,y₃).

Then the equation of the tangent at the point Q(x₂,y₂) is

yy₂ = 2a(x+x₂)

P lies in this line so

** y₁y₂ = 2a(x₁+x₂)**.

Similarly the equation of the tangent at the point R(x₃,y₃) is

yy₃ = 2a(x+x₃)

P lies in this line so,

** y₁y₃ = 2a(x₁+x₃).**

From the above two equations the points of contact Q and R lie on the line yy₁=2a(x+x₁).

So the equation of chord of contact is **yy₁=2a(x+x₁).**

**Example:**

Find the equation of the chord of contact of tangents from the point (0,1) for the parabola y² = x-1.

Solution:

The equation of the given parabola is y² = x-1.

Equation of the parabola with the change of origin is Y²= X-1

a = 1/4.

Equation of chord of contact to the parabola at the point (x₁,y₁) is

yy₁=2a(x+x₁)

y(1) = 2(1/4)(x-1-1)

y = 1/2 (x-2)

2y = x-2

** x-2y-2=0**. is the required equation.

**Example: **

Find the equation of the chord of contact of tangents from the point (-2,1) to the parabola x² =y.

Solution:

Equation of the parabola is x²=y.

4a=1

a =1/4.

Equation of chord of contact to the parabola at the point (x₁,y₁) is

xx₁=2a(y+y₁)

Here (x₁,y₁) is (-2,1).

So equation of chord of contact is

x(-2) = 2(1/4)(y+1)

-2x = (1/2)y +1/2

-4x = y+1

**4x+y+1=0** is the required equation.

Parents and teachers can guide the students to solve the examples given in this page. Students can solve the examples following the same method. If you have any doubt please contact us, we will help you to solve your doubts.

[?]Subscribe To This Site