Question

Find the equation of the locus of a point, the tangents from which to the parabola y² = 18x are such that some of their slopes is –3

Answer 1

For y² = 18x, 4a = 18

∴ a = `9/2`

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

Let m be the slope of the tangent drawn from P to the parabola.

∴ its equation is y = `"mx" + "a"/"m"`

i.e., y = `"mx" + 9/(2"m")`     ...(1)

Since it passes through P(x₁, y₁), we get,

y₁ = `"mx"_1 + 9/(2"m")`

= `(2"x"_1"m"^2 + 9)/(2"m")`

∴ 2x₁m² – 2y₁m + 9 = 0

This is a quadratic equation in m

Its roots m₁ and m₂ are the slopes of the tangents drawn from P.

∴ m₁ + m₂ = `-((-2"y"_1)/(2"x"_1)) = "y"_1/"x"_1`

But it is given that m₁ + m₂ = – 3

∴ `"y"_1/"x"_1`= – 3

∴ 3x₁ + y₁ = 0

Replacing x₁ by x and y₁ by y, the equation of the required locus is 3x + y = 0.