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 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

(yy1)=y12a(xx1)(y-y_{1})= \frac{-y_{1}}{2a}(x-x_{1})

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

Equation of Parabola

Normal at (x1, y1)

y2 = 4ax

(y – y1) = (-y1/2a)(x – x1)

y2 = -4ax

(y – y1) = (y1/2a)(x – x1)

x2 = 4ay

(y – y1) = (-2a/x1)(x – x1)

x2 = -4ay

(y – y1) = (2a/x1)(x – x1)

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 Parabola

Point of Contact

Equation of Normal

y2 = 4ax

(am2, -2am)

y = mx – 2am – am3

y2 = -4ax

(-am2, 2am)

y = mx + 2am + am3

x2 = 4ay

(-2a/m, a/m2)

y = mx + 2a + a/m2

x2 = -4ay

(2a/m, -a/m2)

y = mx – 2a – a/m2

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 normal of standard parabolas at t.

Equation

Parametric Coordinates

Normal at t

y2 = 4ax

(at2, 2at)

y = – tx + 2at + at3

y2 = -4ax

(-at2, 2at)

y = tx + 2at + at3

x2 = 4ay

(2at, at2)

x = – ty + 2at + at3

x2 = -4ay

(2at, -at2)

x = ty + 2at + at3

Solved Examples

Let us have a look at some examples on the equation of normal to 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 parabola is y2 = 8x

Comparing with standard equation,

4ax = 8x

So a = 8/4 = 2

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

Required equation of 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)

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:

Equation of 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(at12, 2at1’) and Q’(at22, 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

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).

Option c is the answer.

Also Read:

Previous year questions on Parabola

Tangents and Normals

Related Video links

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