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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
Solution
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.
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