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प्रश्न
Answer the following:
Two tangents to the hyperbola `x^2/"a"^2 - y^2/"b"^2` = 1 make angles θ1, θ2, with the transverse axis. Find the locus of their point of intersection if tan θ1 + tan θ2 = k
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उत्तर
Given equation of the hyperbola is `x^2/"a"^2 - y^2/"b"^2` = 1.
Let θ1 and θ2 be the inclinations.
m1 = tan θ1, m2 = tan θ2
Let P(x1, y1) be a point on the hyperbola
Equation of a tangent with slope ‘m’ to the hyperbola
`x^2/"a"^2 - y^2/"b"^2` = 1 is y = `"m"x ± sqrt("a"^2"m"^2 - "b"^2)`
This tangent passes through P(x1, y1).
∴ y1 = `"m"x_1 ± sqrt("a"^2"m"^2 - "b"^2)`
∴ (y1 – mx1)2 = a2m2 – b2
∴ (`"x"_1^2` – a2)m2 – 2x1y1m + (`"y"_1^2` + b2) = 0 …(i)
This is a quadratic equation in ‘m’.
It has two roots say m1 and m2, which are the slopes of two tangents drawn from P.
∴ m1 + m2 = `(2"x"_1"y"_1)/("x"_1^2-"a"^2)`
Since tan θ1 + tan θ2 = k,
`(2"x"_1"y"_1)/("x"_1^2-"a"^2)` = k
∴ P(x1, y1) moves on the curve whose equation is k(x2 – a2) = 2xy.
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