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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 :

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

[tex]4=3c\implies c=\dfrac43[/tex]

So when [tex]t=2[/tex], the [tex]x[/tex]-coordinate of the point [tex]p[/tex] would be

[tex]x=\dfrac43(2)=\dfrac83[/tex]

I'm not sure how [tex]y[/tex] is parameterized, but my best guess is that you mean to say

[tex]y(t)=\dfrac{c^2}8t^2[/tex]

which means when [tex]t=2[/tex] we would get a [tex]y[/tex]-coordinate of

[tex]y=\dfrac{\left(\frac43\right)^2}82^2=\dfrac89[/tex]

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