For what values of a and b is the line 3x y = b tangent to the parabola y = ax2 when x = 2?

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

Condition for Tangency

one. The line y = mx + c is a tangent to the parabola y^two = 4ax, if c = a/yard.

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

3. The line ten 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 betoken P(x_one, y₁) is yy₁ = 2a(x + 10₁).

Proof:

Consider the parabola y² = 4ax….(i)

Let the point (10_one, y₁) lie on information technology.

So we can write y₁ ² = 4ax_i …(ii)

Differentiate equation (i) with respect to x.

2y (dy/dx) = 4a

=> dy/dx = 4a/2y

= 2a/y (gradient)

Let m be the gradient of the tangent at point (x_ane, y₁) then

(dy/dx) at (10_i, y_ane) = m = 2a/y₁

The equation of tangent at (10₁, y₁) is given by

(y – y₁) = m(ten – x₁)

=> (y – y₁) = (2a/y_ane)(x – x_one)

=> yy₁ – y₁ ^two = 2a(x – x_one)

Substitute (ii) in above equation

yy₁ – 4ax₁ = 2a(10 – ten₁)

=> yy₁ = 2ax – 2ax₁ + 4ax_one

=> yy₁ = 2ax + 2ax₁

=> yy₁ = 2a(x + 10₁) ….(iii) required equation.

2. Slope form:

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

The point of contact is (a/m², 2a/m).

Proof:

Allow yⁱⁱ = 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 a 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 xⁱⁱ = 4ay, and then the tangent is given by y = mx – am². The indicate of contact is (2am, am^two)

iii. Parametric form:

The equation of the tangent to the parabola yⁱⁱ = 4ax at (at², 2at) is ty = 10 + at².

Proof:

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

Equation of the tangent at P on the parabola is

(Substitute x_one = at^two and y₁ = 2at in (iii) of proof given in bespeak form.)

=> y(2at) = 2a(ten + atⁱⁱ)

=> yt = (ten + at^two) which is the required equation.

Besides Read

Conic Sections

JEE Previous Twelvemonth Questions with Solutions on Parabola

Solved Examples

Case one:

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

a. 10 – 2y + 8 = 0

b. x – 2x – 8 = 0

c. x + 2y + eight = 0

d. none of these

Solution:

We utilize a parametric class equation.

Given y² = 8x

=> a = 2

t = two

(at², 2at) = (8, viii)

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

=> 2y = 10 + viii

=> ten – 2y + viii = 0

Hence option (a) is the answer.

Instance 2:

The status 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 equation of line ten cos θ + y sin θ = P

=> y sin θ = – ten cos θ + P

=> y = (-cot θ)ten + P/sin θ ….(1)

Comparing above equation with y = mx + c

Nosotros get chiliad = -cot θ, c = P/sin θ

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

=> P/sin θ = a/-cot θ

=> P = -a sinⁱⁱθ/cos θ

=> P cos θ + a sin^twoθ = 0, which is the required equation.

Hence, option (a) is the respond.

Related video

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Source: https://byjus.com/jee/tangent-to-a-parabola/