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Question
Answer the following:
The slopes of the tangents drawn from P to the parabola y2 = 4ax are m1 and m2, show that m1 − m2 = k, where k is a constant.
Solution
Let P(x1, y1) be any point on the parabola y2 = 4ax
Equation of tangent to the parabola y2 = 4ax having slope m is y = `"m"x + "a"/"m"`
This tangent passes through P(x1, y1)
∴ y1 = `"m"x_1 + "a"/"m"`
∴ my1 = m2x1 + a
∴ m2x1 – my1 + a = 0
This is a quadratic equation in ‘m’.
The roots m1 and m2 of this quadratic equation are the slopes of the tangents drawn from P.
∴ m1 + m2 = `y_1/x_1`, m1·m2 = `"a"/x_1`
(m1 – m2)2 = (m1 + m2)2 – 4m1m2
= `(y_1/x_1)^2 - (4"a")/x_1`
= `(y_1^2 - 4"a"x_1)/x_1^2`
∴ m1 – m2 = `sqrt((y_1^2 - 4"a"x_1)/x_1^2)`
Since (x1, y1) and a are constants, m1 − m2 is a constant.
∴ m1 – m2 = k, where k is constant.
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