Question

If the constant c is chosen so that the curve given parametrically by ct, c 2 8 t 2 , 0 ≤ t ≤ 3 , is the arc of the parabola 8y = x 2 from (0, 0) to (4, 2), find the coordinates of the point p on this arc corresponding to t = 2.

Answer 1

Since the parabola is given by $8y=x^2$, and $x$ is parameterized by $x(t)=ct$, it follows that when $t=3$ and $x=4$, we have

$4=3c\implies c=\dfrac43$

So when $t=2$, the $x$-coordinate of the point $p$ would be

$x=\dfrac43(2)=\dfrac83$

I'm not sure how $y$ is parameterized, but my best guess is that you mean to say

$y(t)=\dfrac{c^2}8t^2$

which means when $t=2$ we would get a $y$-coordinate of

$y=\dfrac{\left(\frac43\right)^2}82^2=\dfrac89$