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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
Solution
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.
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