Equations of Normal to a Parabola

The normal to a parabola is perpendicular to the tangent of the parabola. Also, it will pass through the point of intersection of the tangent to the parabola. In this article, we will learn the different equations of normal to a parabola and examples on how to find the equation of normal to a parabola. Equations of normal to a parabola is an important topic for the JEE exam. Students can definitely expect questions from this topic.

Consider a parabola y² = 4ax. Let T be a tangent to this parabola at point P. Here, P is the point of contact for the tangent T. The normal will be a line passing through P and perpendicular to T. So, we can say that the line perpendicular to the tangent at the point of contact is called the normal to the parabola at that point. Note that the product of the slope of the tangent and the normal to the parabola is -1. The various equations of normal to a parabola are given below.

The equations are given in point form, parametric form and slope form.

Equation of Normal in Point Form

The equation of normal to the parabola y² = 4ax at point P(x₁, y₁) is

The following table gives the equation of the normal of standard parabolas at (x₁, y₁) in point form.

Equation of Normal in Slope Form

The equation of normal to the parabola y² = 4ax of slope m is given by the equation

y = mx – 2am – am³. Here, the point of contact is (am², -2am).

The table given below gives the equation of normal, point of contact, and criteria for normality in terms of slope m.

Equation of Normal in Parametric Form

The equation of normal to the parabola y² = 4ax at P(t) = P(at², 2at) is given by

y = -tx + 2at + at³.

The table given below gives the equation of the normal of standard parabolas at t.

Solved Examples

Let us have a look at some examples on the equation of normal to a parabola.

Example 1:

P and Q are the points t₁ and t₂ on the parabola y² = 4ax. If the normals to the parabola at P and Q meet at R (a point on the parabola), then the value of t₁t₂ is:

a) 1

b) 2

c) ½

d) None of these

Solution:

Consider the normals at P and Q meet at R(at², 2at).

So, t = – t₁ – 2/t₁ and t = – t₂ – 2/t₂

Hence, t₁ + 2/t₁ = t₂ + 2/t₂

(t₁ – t₂) = 2(t₁ – t₂)/t₁ t₂

So, t₁t₂ = 2

Hence, option b is the answer.

Example 2:

Find the equations of the tangent and normal to the parabola y² = 8x at (2, 4).

Solution:

Given equation of a parabola is y² = 8x

On comparing with the standard equation,

4ax = 8x

So, a = 8/4 = 2

Equation of tangent at (x₁, y₁) to y² = 4ax is yy₁ = 2a(x + x₁)

The required equation of the tangent at (2, 4) to y² = 8x is 4y = 2× 2(x + 2)

=> 4y = 4(x + 2)

=> y = x + 2

Equation of normal at (x₁, y₁) is (y – y₁) = (-y₁/2a)(x – x₁)

The required equation of normal is

=> y – 4 = (-4/4)(x – 2)

=> y – 4 + x – 2 = 0

=> x + y – 6 = 0

Example 3:

If the normals to the parabola y² = 4ax at the end of its latus rectum meets the parabola at Q and Q’, then show that QQ’ = 12a.

Solution:

The equation of the given parabola is y² = 4ax

Ends of the latus rectum are P(a, 2a) and P’(a, -2a)

Point P has parameter t₁ = 1 and P’ has parameter t₂ = -1

Let Q(at₁’², 2at₁’) and Q’(at₂’², 2at₂’)

Normal at P meets Q whose parameter t₁’ = -t₁ – 2/t₁

= -1 – 2 (Put t₁ = 1)

= -3

Normal at P’ meets Q’ whose parameter t₂’ = -t₂ – 2/t₂

= 1 + 2 (Put t₂ = -1)

= 3

Coordinate of Q = (9a, -6a)

Coordinate of Q’ = (9a, 6a)

QQ’ = √(6a + 6a)² + 0 (distance formula)

= 12a

Example 4:

Find the equation of normal to the parabola x² = 4y at (6, 9).

Solution:

Given parabola is x² = 4y

On comparing with the standard equation, we get, a = 1

The equation of normal to the parabola x² = 4ay at (x₁, y₁) is

(y – y₁) = (-2a/x₁)(x – x₁)

Required equation is y – 9 = (-2/6)(x – 6)

-3y + 27 = x – 6

=> x + 3y – 33 = 0

Example 5:

If the line x + y = 1 touches the parabola y² – y +x = 0, then the coordinates of the point of contact are:

a) (1,1)

b) (1, 0)

c) (0, 1)

d) None of these

Solution:

Given line is x + y = 1

=> x = -y + 1 …(i)

Given line touches the parabola y² – y +x = 0…(ii)

Put (i) in (ii)

y² – y – y + 1 = 0

=> y² – 2y + 1 = 0

=> (y – 1)² = 0

=> y = 1

So, x = 0

Hence, the point of contact is (0, 1).

Hence, option c is the answer.

Also Read:

Previous Years’ Questions on Parabola

Tangents and Normals

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