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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 y2 = 4ax in two points real, coincident or imaginary according as a >=< cm condition of tangency is,

 Condition of tangency is  c = am

Note : Line y = mx + c will be tangent to parabola x2 = 4ay if c = -am2

(b) Length of the chord intercepted by the parabola y2 = 4ax on the line y = mx + c is 4m2a(1+m2)(amc)

Note : Length of the focal chord making an angle α with the x-axis is 4a cosec2α.

Equation of Tangent to Parabola y2=4ax

(a) Point form :

The equation of tangent to the given parabola at its point (x1,y1) is

yy1 = 2a(x + x1)

Example : Find the tangent to the parabola y2=16x at (5, 2).

Solution : We have, y2=16x

Compare given equation with y2=4ax

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 + am, (m 0)

The point of contact is (am2, 2am)

Example : Find the tangent to the parabola y2=8x whose slope is 3.

Solution : We have, y2=8x

Compare given equation with y2=4ax

a = 2

Hence, required equation of tangent is y = 3x + 23

(c) Parametric form :

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

ty = x + at2

Note – Point of intersection of the tangents at the points t1 & t2 is [at1t2, a(t1+t2)].

Example : Find the equation of the tangents to the parabola y2 = 9x which go through the point (4,10).

Solution : tangent to the parabola y2 = 9x is

y = mx + 94m

Since it passes through (4,10)

10 = 4m + 94m 16m2 – 40m + 9 = 0

m = 14, 94

Equation of tangent’s are y = x4 + 9 & y = 9x4 + 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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