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Question
Find the equation of tangent to the parabola y2 = 12x from the point (2, 5)
Solution
Given equation of the parabola is y2 = 12x.
Comparing this equation with y2 = 4ax, we get
4a = 12
∴ a = 3
Equation of tangent to the parabola y2 = 4ax having slope m is y = `"mx" + "a"/"m"`
Since the tangent passes through the point (2, 5),
5 = `2"m" + 3/"m"`
∴ 5m = 2m2 + 3
∴ 2m2 – 5m + 3 = 0
∴ 2m2 – 2m – 3m + 3 = 0
∴ 2m(m – 1) – 3(m – 1) = 0
∴ (m – 1)(2m – 3) = 0
∴ m = 1 or m = `3/2`
These are the slopes of the required tangents.
By slope point form, y – y1 = m(x – x1), the equations of the tangents are
y – 5 = 1(x – 2) and y – 5 = `3/2("x" - 2)`
∴ y – 5 = x – 2 and 2y – 10 = 3x – 6
∴ x – y + 3 = 0 and 3x – 2y + 4 = 0
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