Tangent to a Parabola

A line that touches the parabola exactly at one point is called the tangent to a parabola. In this article, we learn the equation of the tangent to a parabola and the point of contact of the tangent to a parabola, along with solved examples.

Condition for Tangency

1. The line y = mx + c is a tangent to the parabola y² = 4ax, if c = a/m.

2. The line y = mx + c is a tangent to the parabola x² = 4ay, if c = -am².

3. The line x cos θ + y sin θ = p is a tangent to the parabola y² = 4ax, if a sin²θ + p cos θ = 0

Equation of Tangent to a Parabola

1. Point Form

The equation of tangent to the parabola y² = 4ax at point P(x₁, y₁) is yy₁ = 2a(x + x₁).

Proof:

Consider the parabola y² = 4ax….(i)

Let the point (x₁, y₁) lie on it.

So, we can write y₁² = 4ax₁ …(ii)

Differentiate equation (i) with respect to x.

2y (dy/dx) = 4a

=> dy/dx = 4a/2y

= 2a/y (slope)

Let m be the slope of the tangent at point (x₁, y₁), then

(dy/dx) at (x₁, y₁) = m = 2a/y₁

The equation of tangent at (x₁, y₁) is given by

(y – y₁) = m(x – x₁)

=> (y – y₁) = (2a/y₁)(x – x₁)

=> yy₁ – y₁² = 2a(x – x₁)

Substitute (ii) in the above equation

yy₁ – 4ax₁ = 2a(x – x₁)

=> yy₁ = 2ax – 2ax₁ + 4ax₁

=> yy₁ = 2ax + 2ax₁

=> yy₁ = 2a(x + x₁) ….(iii) required equation.

2. Slope form:

a. The equation of the tangent of the parabola y² = 4ax is y = mx + a/m, where c = a/m.

The point of contact is (a/m², 2a/m).

Proof:

Let y² = 4ax be the parabola. Suppose the line y = mx + c is the tangent to the parabola.

The condition that the line y = mx + c is the tangent to the parabola y² = 4ax is c = a/m.

Put c = a/m in y = mx + c.

Here, m is the slope of the tangent.

=> y = mx + a/m, which is the required equation.

b. If the parabola is given by x² = 4ay, then the tangent is given by y = mx – am². The point of contact is (2am, am²)

3. Parametric form:

The equation of the tangent to the parabola y² = 4ax at (at², 2at) is ty = x + at².

Proof:

Let P(at², 2at) be the point on the parabola through which the tangent passes.

The equation of the tangent at P on the parabola is

(Substitute x₁ = at² and y₁ = 2at in (iii) of proof given in point form.)

=> y(2at) = 2a(x + at²)

=> yt = (x + at²), which is the required equation.

Also Read

Conic Sections

JEE Previous Year Questions with Solutions on Parabola

Solved Examples

Example 1:

The equation of the tangent to the parabola y² = 8x at t = 2 is

a. x – 2y + 8 = 0

b. x – 2x – 8 = 0

c. x + 2y + 8 = 0

d. none of these

Solution:

We use a parametric form equation.

Given y² = 8x

=> a = 2

t = 2

(at², 2at) = (8, 8)

The equation of the tangent to the parabola y² = 4ax at (at², 2at) is ty = x + at².

=> 2y = x + 8

=> x – 2y + 8 = 0

Hence, option (a) is the answer.

Example 2:

The condition that the line x cos θ + y sin θ = P touches the parabola y² = 4ax is

a. P cos θ + a sin²θ = 0

b. cos θ + a P sin²θ = 0

c. P cos θ – a sin²θ = 0

d. None of these

Solution:

Given, the equation of line x cos θ + y sin θ = P

=> y sin θ = – x cos θ + P

=> y = (-cot θ)x + P/sin θ ….(1)

Comparing the above equation with y = mx + c

We get m = -cot θ, c = P/sin θ

From the condition of tangency, c = a/m.

=> P/sin θ = a/-cot θ

=> P = -a sin²θ/cos θ

=> P cos θ + a sin²θ = 0, which is the required equation.

Hence, option (a) is the answer.

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