The equation of normal to parabola in point form, slope form and parametric form are given below with examples.

## Equation of Normal to Parabola \(y^2 = 4ax\)

### (a) Point form :

The equation of normal to the given parabola at its point **(\(x_1, y_1\))** is

y – \(y_1\) = \(-y_1\over 2a\)(x – \(x_1\))

Example : Find the normal to the parabola \(y^2 = 32x\) at (3, 1).

Solution : We have, \(y^2 = 32x\)

Compare given equation with \(y^2 = 4ax\)

\(\implies\) a = 8

Hence, required equation of normal is y – 1 = \(-1\over 16\)(x – 3)

=> 16y + x = 19

**(b) Slope form** :

The equation of normal to the given parabola whose slope is ‘m’, is

y = mx – 2am – a\(m^3\)

The foot of the normal is (\(am^2\), -2am)

Example : Find the normal to the parabola \(y^2 = 4x\) whose slope is 3.

Solution : We have, \(y^2 = 4x\)

Compare given equation with \(y^2 = 4ax\)

a = 1

Hence, required equation of normal is y = 3x – 33

**(c) Parametric form** :

The normal to the given parabola at its point P(t), is

y + tx = 2at + a\(t^3\)

**Note –** Point of intersection of the normals at the points \(t_1\) & \(t_2\) is

[(a\({t_1}^2 + {t_2}^2 + t_1t_2 + 2\)), -a\(t_1t_2(t_1 + t_2)\)].

Example : If two normals drawn from any point to the parabola \(y^2\) = 4ax make angle \(\alpha\) and \(\beta\) with the axis such that \(tan\alpha\).\(tan\beta\) = 2, then find the locus of this point.

Solution : Let the point is (h, k). The equation of any normal to the parabola \(y^2\) = 4ax is

y = mx – 2am – a\(m^3\)

Since it passes through (h,k)

k = mh – 2am – a\(m^3\)

a\(m^3\) + m(2a – h) + k = 0 ……(i)

\(m_1\), \(m_2\) and \(m_3\) are roots of equation, then

\(m_1\).\(m_2\).\(m_3\) = \(-k\over a\)

but \(m_1\).\(m_2\) = 2, \(m_3\) = \(-k\over 2a\)

\(m_3\) is the root of (i)

\(\therefore\) \(k^2\) = 4ah

Thus locus is \(y^2\) = 4ax.

Hope you learnt equation of normal to parabola in point form, slope form and parametric form, learn more concepts of parabola and practice more questions to get ahead in the competition. Good luck!