Given $y=3x^2 - 2$ , why/how is $x= sqrt(3)*t$ and $y = t^2 -2$ the parametric equations?

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Answer 1 · 2018-06-28T12:37:37

Given $y=3x^2-2$

Lets use parametrization given by

$x=t$

Then $y=3t^2-2$ is the expresion given by this paramerization

In order to remove 3 from this expresion we can use instead

$x=t/sqrt3$ in this case

$y=3(t/sqrt3)^2-2=t^2-2$

The parametrization given above result $x=sqrt3t$ then

$y=9t^2-2$

So I think there is a mistake in parametrization of $x$ in asked question. Hope this helps

Given y=3x^2 - 2y=3x^2 - 2, why/how is x= sqrt(3)*tx= sqrt(3)*t and y = t^2 -2y = t^2 -2 the parametric equations?