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Question
For the parabola 3y2 = 16x, find the parameter of the point (27, –12).
Solution
Given equation of the parabola is 3y2 = 16x.
∴ y2 = `16/3"x"`
Comparing this equation with y2 = 4ax, we get,
4a = `16/3`
∴ a = `4/3`
If t is the parameter of the point P on the parabola, then
P(t) ≡ (at2, 2at),
i.e., x = at2 and y = 2at ...(i)
Given point is (27, –12)
Substituting x = 27, y = –12 and a = `4/3` in (i), we get,
27 = `4/3`t2 and –12 = `2(4/3)`t
∴ t2 = `81/4` and t = `(-9)/2`
∴ t = `± 9/2` and t = `(-9)/2`
∴ t = `(-9)/2`
∴ The parameter of the given point is `(-9)/2`.
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