Normal to a Parabola

Normal at the point (x₁, y₁) 

The equation of the tangent at the point (x₁, y₁) is yy₁ = 2a(x + x₁). Since the slope of the tangent = 2a/y₁, 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 m 

In equation (i), put -y₁/2a = m so that y₁ = –2a and x₁ = (y₁²)/4a = am², then the equation becomes y = mx – 2am – am³ 

where m is a parameter. Equation (ii) is the normal at the point (am², –2am) of theparabola. 


Note: 

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 m₁+ m₂+ m₃ = 0, m₁ m₂ + m₂ m₃ + m₃ m₁ = (2a – h)/a, 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.

Illustration: 

          Find the locus of the point of intersection of two normals to a parabolawhich 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 m_3. Let the perpendicularnormals correspond to the values of m₁ and m₂ so that m₁ m₂ = –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).



Normal at the point t 

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

Note: 

            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 (at₁², 2at₁) and (at₂t², 2at₂) is (2a+a(t₁²+t₂²+t₁t₂), a t₁t₂(t₁+t₂)). 

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 = ¾ and slope of QS = – ¾. 

⇒ ∠PSQ = right angle. Hence the result.