Question

Answer the following :

Tangents to the circle x² + y² = a² with inclinations, θ₁ and θ₂ intersect in P. Find the locus of such that cot θ₁ + cot θ₂ = 5

Answer 1

Let P(x₁, y₁) be a point on the required locus. Equations of the tangents to the circle 

x² + y² = a² with slope m are

y = `"mx"  ± sqrt("a"^2(1 + "m"^2)`

Since, these tangents pass through (x₁, y₁).

y₁ = `"mx"_1  ± sqrt("a"^2(1 + "m"^2)`

∴ y₁ – mx₁ = `± sqrt("a"^2(1 + "m"^2)`

∴ `"y"_1^2` – 2mx₁y₁ + `"m"^2"x"_1^2` = a² + a²m²

∴ (`"x"_1^2` – a²)m² – 2mx₁y₁ + (`"y"_1^2` – a²) = 0

This is a quadratic equation which has two roots m₁ and m₂.

∴ m₁ + m₂ = `(2"x"_1"y"_1)/("x"_1^2 - "a"^2) and"m"_1"m"_2 = ("y"_1^2 - "a"^2)/("x"_1^2 - "a"^2)`

Given, cot θ₁ + cot θ₂ = 5

∴ `1/tantheta_1 + 1/tantheta_2` = 5

∴ `1/"m"_1 + 1/"m"_2` = 5

∴ `("m"_1 + "m"_2)/("m"_1"m"_2)` = 5

∴ `((2"x"_1"y"_1)/("x"_1^2 - "a"^2))/(("y"_1^2 - "a"^2)/("x"_1^2 - "a"^2))` = 5

∴  `(2"x"_1"y"_1)/("y"_1^2 - "a"^2)` = 5

∴ 2x₁y₁ = `5"y"_1^2` – 5a²

∴ `5"y"_1^2-2"x"_1"y"_1=5"a"^2`

∴ Equation of the locus of point P is 5y² – 2xy = 5a².