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. 


  • 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


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:


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


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).

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.


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.


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


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 .


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. 


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.


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. 


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.

Related Resources

To read more, Buy study materials of Parabola comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics here.


Upto 50% Scholarship on Live Classes

Course Features

  • Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution