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Question
For the parabola 3y2 = 16x, find the parameter of the point (3, – 4).
Solution
Given the 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 (3, – 4)
Substituting x = 3, y = – 4 and a = `4/3` in (i), we get
3 = `4/3"t"^2` and −4 = `2(4/3)"t"`
∴ t2 = `9/4` and t = `(-3)/2`
∴ t = `±3/2` and t = `(-3)/2`
∴ t = `-3/2`
∴ The parameter of the given point is `-3/2`.
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