Answer
$a)y= \pm \dfrac{x^{3/2}}{8}-1; b) y=\pm \dfrac{\sqrt {1-x^2}}{x}$
Work Step by Step
(a) Here, we have $x=4t^2$
This gives: $t=\pm \dfrac{\sqrt x}{2}$
Then $ y=t^3-1$
or, $y=(\pm \dfrac{\sqrt x}{2})^3-1=\pm \dfrac{x^{(3/2)}}{8}-1$
(b) Here, we have $\dfrac{1}{x}=\sec t$
or, $\sec^2 t=1+\tan^2 t$
This gives: $ \dfrac{1}{x^2}-1=y^2$
Thus, $y=\pm \dfrac{\sqrt {1-x^2}}{x}$