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

Jun 28, 2018

Given $y = 3 {x}^{2} - 2$

Lets use parametrization given by

$x = t$

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

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

$x = \frac{t}{\sqrt{3}}$ in this case

$y = 3 {\left(\frac{t}{\sqrt{3}}\right)}^{2} - 2 = {t}^{2} - 2$

The parametrization given above result $x = \sqrt{3} t$ then

$y = 9 {t}^{2} - 2$

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