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# Normal to a Parabola

## What do we mean by a normal to a parabola? The concept of normal is closely related to that of tangents. The product of the slopes of tangent and normal is equal to -1. This concept is extremely important and it often fetches questiosn in competitive exams like the JEE. The normals to a parabola can be written in various forms, i.e. there can be various equations of normals to a parabola. We discuss the various forms one by one:

Normal at the point (x1, y1

The equation of the tangent at the point (x1, y1) is yy1 = 2a(x + x1).If the slope of the tangent = 2a/y1 then the slope of the normal is –y1/2a. Also it passes through (x1, y1).

Hence, its equation is y – y1 = -y1/2a (x – x1). … (i)

Normal in terms of slope (m)

In equation (i), put -y1/2a = m so that y1 = –2a and x1 = (y12)/4a = am2, then the equation becomes y = mx – 2am – amwhere m is a parameter. Equation (ii) is the normal at the point (am2, –2am) of the parabola.

Remark:

• If this normal passes through a point (h, k), then k = mh – 2am – am3.

• For a given parabola and a given point (h, k), this cubic in m has three roots say m1, m2, m3 i.e. from (h, k) three normals can be drawn to the parabola whose slopes are m1, m2, m3. For this cubic, we have m+ m+ m3 = 0, m1 m2 + m2 m3 + m3 m1 = (2a – h)/a and m1 m2 m3 = –k/a.

• If we have an extra condition about the normals drawn from a point (h, k) to a given parabola y2 = 4ax then by eliminating m1, m2, m3 from these four relations between m1, m2, m3, we can get the locus of (h, k).

• Since the sum of the roots is equal to zero, the sum of the ordinates of the feet of the normals from a given point is zero. These points are called Co-Normal Points.

Normal at the point t

The normal, being perpendicular to the tangent at (at, 2at) is given by y = –tx + 2at + at3

Remark:

If normal at the point at1 meets the parabola again at the point at2, then at2 = – at1 – 2/at1 .

Point of intersection of the normals to the parabola y2 = 4ax t1 and t2 are (2a + a(t1+ t2+ t1t2), -a t1t2(t1+t2)).

## Properties of a normal to a Parabola y2 = 4ax

• From a given point, a total of three normals can be drawn to the parabola y2 = 4ax.

• The algabraic sum of the slope of these normals is zero.

• The algebraic sum of the ordinates of their feet is zero.

• The centroid of the triangle formed by joining their feet lies on the axis of the parabola.

• The tangent at one extremity of a focal chord of a parabola is parallel to the normal at the other extremity.

• The normals lying on the extremities of the latus rectum of a parabola intersect at right angles on the xais of the parabola.

• Atleast one of the normal is real since imaginary normals always occur in pairs.

• The length of the normal is gievn by the formula a(t- t2)(t+ t2)2 + 4 = 4a(t12 + 1)3/2 /t12

• When the feet of the normals of the parabola are joined, they form a triangle whose centroid is given by the formula ((am12 + am22 + am32)/3, (2am1 + 2am2 + 2am3)/ 3) which is equal to (am12 + am22 + am32) /3, 0)

We now discuss some of the illustrations based on the above concepts:

Illustration:

Find the locus of the point of intersection of two normals to a parabola which are at right angles to one another.

Solution:

The equation of the normal to the parabola y2 = 4ax is y = mx – 2am – am3.

It passes through the point (h, k) if

k = mh – 2am – am3 => am3 + m(2a – h) + k = 0. … (1)

Let the roots of the above equation be m1, m2and m3. Let the perpendicular normals correspond to the values of m1 and m2 so that m1m2 = –1.

From equation (1), m1 m2 m3 = -k/a. Since m1 m2 = –1, m3 = k/a.

Since m3 is a root of (1), we have a (k/a)3+k/a (2a – h) + k = 0. ⇒ k2 + a(2a – h) + a2 = 0

⇒ k2 = a(h – 3a).

Hence the locus of (h, k) is y2 = a(x – 3a).

Illustration:
Prove that the normal chord to a parabola at the point whose ordinate is equal to the abscissa subtends a right angle at the focus.

Solution:

If the normal to the parabola y2 = 4ax at P(at1t2, 2at1) meets it again at the point t2, then we have t2 = – t1 – 2/t1

If the abscissa and the ordinates of P be equal, then at12 = 2at1

⇒ t1 = 2 (rejecting t1 = 0) ? t2 = – 2 – 1 = – 3

The co-ordinates of P and Q are therefore (4a, 4a) and (9a, – 6a) respectively.

The focus is the point S (a, 0).

Slope of PS = 3/4 and slope of QS = – 3/4.

⇒ ∠PSQ = right angle. Hence the result.

Illustration:

Find the locus of the middle points of the normal chords of the parabola y2 = 4ax.

Solution:

Equation of the normal chord at any point (at2, 2at) of the parabola is

y + tx = 2at + at3. … (1)

Equation of the chord with mid point (x1, y1) is T = S1

or yy1 – 2a(x + x1) = y12 – 4ax1 or yy1 – 2ax = y12 – 2ax1. … (2)

Since equations (1) and (2) are identical, 1/y1 = t /(-2a) = (2at + at3)/t = 2a + ((-2a)/y1)2

or -(y12) /2a + x1 = 2a + 4a/(y12) or x1 – 2a = (y12) /2a + 4a/(y12)

Hence the locus of the middle point (x1, y1) is x – 2a = y2/2a + 4a/y2 .

Illustration:

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), show that t1t2 = 2.

Solution:

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

Then t = – t1 – 2/t1 and t = – t2 – 2/t2 .

Therefore t1 + 2/t1 = t2 + 2/t2 ⇒ (t1 – t2) = 2(t1-t2 )/(t1 t2 ) ⇒ t1t2 = 2.

Illustration:

Find the equations of the normals to the parabola y2 = 4ax at the extremities of its latus rectum. If the normals meet the parabola, again at P and Q, prove that PQ = 12a.

Solution:

The ends of the latus rectum are (a, 2a) and (a, –2a). The equations of the normals to the parabola at these points are (put t = 1 and –1)

y + x = 3a and y – x = 3a.

These lines meet the parabola again at P(9a, 6a) and Q(9a, 6a) respectively.

⇒ PQ = 6a + 6a = 12a.

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