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Question
Find the equation of common tangent to the parabola y2 = 4x and x2 = 32y
Solution
Given equation of the parabola is y2 = 4x.
Comparing this equation with y2 = 4ax, we get
4a = 4
∴ a = 1
Let the equation of common tangent be
y = `"mx" + 1/"m"` ...(i)
Substituting y = `"mx" + 1/"m"` in x2 = 32y, we get
x2 = `32("mx" + 1/"m")`
= `32 "mx" + 32/"m"`
∴ mx2 = 32m2x + 32
∴ mx2 – 32m2x – 32 = 0 …(ii)
Line (i) touches the parabola x2 = 32y.
∴ The quadratic equation (ii) in x has equal roots.
∴ Discriminant = 0
∴ (– 32m2)2 – 4(m) (– 32) = 0
∴ 1024m4 + 128m = 0
∴ 128m (8m3 + 1) = 0
∴ 8m3 + 1 = 0 ...[∵ m ≠ 0]
∴ m3 = `-1/8`
∴ m = `-1/2`
Substituting m = `-1/2` in (i), we get
y = `-1/2"x" + 1/((-1/2)`
∴ y = `-1/2 "x" - 2`
∴ x + 2y + 4 = 0, which is the equation of the common tangent.
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