Question

Vik. WW 10nws yucu
(i) For the parabola 3y2 = 16x, find the parameter of the point
(3, 4)​

Answer 1

3y² = 16x

y² = $\frac{16}{3}$ x

on comparing with y² = 4ax, we get

4a = $\frac{16}{3}$

a = $\frac{4}{3}$ ... (i)

now, for (x, y) = (3, 4)

considering the parametric forms, we get;

at² = 3 ... (x = at²)

$\frac{4}{3}$ t² = 3 ... (from (i))

t² = 3 $(\frac{3}{4})$

t² = $\frac{9}{4}$

t = $± \frac{3}{2}$

but,

2at = 4 ... (y = 2at)

at = 2

$\frac{4}{3}$ t = 2 ... (from (i))

t = $\frac{3}{2}$

hence,

t ≠ $- \frac{3}{2}$

Hence, parameter of the point (3, -4) is

$\frac{3}{2}$