Question

For the parabola 3y² = 16x, find the parameter of the point (27, –12).

Answer 1

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`.