Answer
$a)~y= \pm \dfrac{x^{3/2}}{8}-1$ and b) $y=\pm \dfrac{\sqrt {1-x^2}}{x}$
Work Step by Step
a) Here, $x=4t^2 \implies t=\pm \dfrac{\sqrt x}{2}$
Then $ y=t^3-1=(\pm \dfrac{\sqrt x}{2})^3-1=\pm \dfrac{x^{3/2}}{8}-1$
b) Here, $\dfrac{1}{x}=\sec t \implies \sec^2 t=\tan^2 t+1$
Then $ \dfrac{1}{x^2}-1=y^2$
This implies that $y=\pm \dfrac{\sqrt {1-x^2}}{x}$
