## Thomas' Calculus 13th Edition

$x=\dfrac{at}{\sqrt{1+t^2}}$; $y=\dfrac{a}{\sqrt{1+t^2}}$ ; $-\infty \lt t \lt \infty$
Here, we have $x^2+y^2=a^2$ This implies that $y=\sqrt {a^2-x^2}$ and slope: $t=\dfrac{dy}{dx}=-\dfrac{x}{\sqrt {a^2-x^2}}$ Now,we have $x^2=\dfrac{(at)^2}{1+t^2} \implies x=\dfrac{at}{\sqrt{1+t^2}}$ Then, $y= \sqrt {a^2-\dfrac{a^2t^2}{1+t^2}}=\dfrac{a}{\sqrt{1+t^2}}$ Hence, $x=\dfrac{at}{\sqrt{1+t^2}}$; $y=\dfrac{a}{\sqrt{1+t^2}}$ ; $-\infty \lt t \lt \infty$