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Can $(a(\sin(t))^2 , 2a \sin(t))$ be the parametric coordinates of the parabola $y^2 = 4ax$ ?

I found that these coordinates satisfy the equation of the parabola but my friend says that although these satisfy the equation , these can not be the coordinates. Can any body give me an explanation for this?

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Because of the limits on the output values of $\sin t$ you will only get part of the parabola, not the whole curve

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The coordinates of the point are periodic, and share a period. They satisfy the equation of the parabola, and they are continuous. You can therefore expect that as $t$ increases, the point will trace out some portion of the parabola over and over.

Specifically (assuming $a>0$), if $t$ increases from $0$, the point will move from $(0,0)$ up and to the right until it reaches $(a,2a)$ when $t=\frac\pi2$. As $t$ continues to increase, the motion will reverse and the point will arrive back at the origin when $t=\pi$. It will continue through the origin and move down and to the right until it reaches $(a,-2a)$ when $t=\frac{3\pi}2$. Then it will reverse and return to the origin, reaching $(0,0)$ when $t=2\pi$. After that, everything repeats. The point continues to swing up and down along the same part of the parabola as $t$ increases without bound.

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If you convert $y^2 = 4ax$ to parametric equation with $y = t$, that will be $[x,y]^T=[\frac{t^2}{4a}, t]^T$

If we let $y = 2at$, then the parametric equation will be $[x,y]^T=[at^2, 2at]^T$. This will just ride faster along the curve by a factor of $2a$ compared to $[x,y]^T=[\frac{t^2}{4a}, t]^T$.

If we let $t = sin(\theta)$, that would limit the parametric variable $t$ to $-1..1$ values only.

Applying it to $[at^2, 2at]^T$, the parametric equation will now be $[x,y]^T=[a\sin^2(\theta), 2a \sin(\theta)]^T$ which is the one in your question.

Therefore, $[a\sin^2(\theta), 2a \sin(\theta)]$ is based from the correct parametric equation of $y^2 = 4ax$, only limited to $x=(0..a)$ and $y=(-2a..2a)$.

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