Equation of Tangent to Parabola in all Forms

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

Condition of Tangency for Parabola :

(a)  The line y = mx + c meets the parabola \(y^2\) = 4ax in two points real, coincident or imaginary according as a >=< cm \(\implies\) condition of tangency is,

 Condition of tangency is  c = \(a\over m\)

Note : Line y = mx + c will be tangent to parabola \(x^2\) = 4ay if c = -a\(m^2\)

(b) Length of the chord intercepted by the parabola \(y^2\) = 4ax on the line y = mx + c is \(4\over m^2\)\(\sqrt{a(1+m^2)(a-mc)}\)

Note : Length of the focal chord making an angle \(\alpha\) with the x-axis is 4a \({cosec}^2\alpha\).

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

(a) Point form :

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

y\(y_1\) = 2a(x + \(x_1\))

Example : Find the tangent to the parabola \(y^2 = 16x\) at (5, 2).

Solution : We have, \(y^2 = 16x\)

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

\(\implies\) a = 4

Hence, required equation of tangent is 2y = 8(x + 5)

= 2y = 8x + 40

(b) Slope form :

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

y = mx + \(a\over m\), (m \(\ne\) 0)

The point of contact is (\(a\over m^2\), \(2a\over m\))

Example : Find the tangent to the parabola \(y^2 = 8x\) whose slope is 3.

Solution : We have, \(y^2 = 8x\)

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

a = 2

Hence, required equation of tangent is y = 3x + \(2\over 3\)

(c) Parametric form :

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

ty = x + a\(t^2\)

Note – Point of intersection of the tangents at the points \(t_1\) & \(t_2\) is [a\(t_1t_2\), a(\(t_1 + t_2\))].

Example : Find the equation of the tangents to the parabola \(y^2\) = 9x which go through the point (4,10).

Solution : tangent to the parabola \(y^2\) = 9x is

y = mx + \(9\over 4m\)

Since it passes through (4,10)

\(\therefore\) 10 = 4m + \(9\over 4m\) \(\implies\) 16\(m^2\) – 40m + 9 = 0

m = \(1\over 4\), \(9\over 4\)

\(\therefore\) Equation of tangent’s are y = \(x\over 4\) + 9 & y = \(9x\over 4\) + 1

Hope you learnt equation of tangent 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!

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