Question

For the parabola 3y² = 16x, find the parameter of the point (3, – 4).

Answer 1

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`.