Question

Two tangents to the parabola y² = 8x meet the tangents at the vertex in the point P and Q. If PQ = 4, prove that the equation of the locus of the point of intersection of two tangent is y² = 8(x + 2).

Answer 1

Given parabola is y² = 8x

Comparing with y² = 4ax, we get,

4a = 8

∴ a = 2

Let M(t₁) and N(t₂) be any two points on the parabola.

The equations of tangents at M and N are

`"yt"_1 = "x" + 2"t"_1^2`   ...(1)

`"yt"_2 = "x" + 2"t"_2^2`  ...(2)  ...[∵ a = 2]

Let tangent at M meet the tangent at the vertex in P.

But tangent at the vertex is Y-axis whose equation is x = 0.

∴ to find P, put x = 0 in (1)

∴ yt₁ = `2"t"_1^2`

∴ y = 2t₁ ...(t₁ ≠ 0 otherwise tangent at M will be x = 0)

∴ P = (0, 2t₁)

Similarly, Q = (0, 2t₂)

It is given that PQ = 4

∴ |2t₁ – 2t₂| = 4

∴ |t₁ – t₂| = 2 ...(3)

Let R = (x₁, y₁) be any point on the required locus

Then R is the point of intersection of tangents at M and N.

To find R, we solve (1) and (2).

Subtracting (2) from (1), we get

y(t₁ – t₂) = `2"t"_1^2-2"t"_2^2` = 2(t₁ – t₂)(t₁ + t₂)

∴ y = 2(t₁ + t₂)  ...[∵ M, N are distinct ∴ t₁ ≠ t₂]

i.e., y₁ = 2(t₁ + t₂) ...(4)

∴ from (1), we get

2t₁(t₁ + t₂) = x + `2"t"_1^2`

∴ 2t₁t₂ = x i.e. x₁ = 2t₁t₂  ...(5)

To find the equation of locus of R(x₁, y₁), we eliminate t₁ and t₂ from the equations (3), (4) and (5).

We know that,

(t₁ + t₂)² = (t₁ − t₂)² + 4t₁t₂

∴ `((y_1)/2)^2 = 4 +  4((x_1)/2)`  ...[By (3), (4) and (5)]

∴ `"y"_1^2` = 16 + 8x₁ = 8(x₁ + 2)

Replacing x₁ by x and y₁ by y, the equation of required locus is y² = 8(x + 2).