## Thomas' Calculus 13th Edition

$a)y= \pm \dfrac{x^{3/2}}{8}-1; b) y=\pm \dfrac{\sqrt {1-x^2}}{x}$
(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, $y=\pm \dfrac{\sqrt {1-x^2}}{x}$