$$\frac{ { x }^{ 2 } }{ 25 } + \frac{ { y }^{ 2 } }{ 9 } =1$$
$x=-\frac{5\sqrt{9-y^{2}}}{3}$
$x=\frac{5\sqrt{9-y^{2}}}{3}$
$y=-\frac{3\sqrt{25-x^{2}}}{5}$
$y=\frac{3\sqrt{25-x^{2}}}{5}$
$x=\frac{5\sqrt{9-y^{2}}}{3}$
$x=-\frac{5\sqrt{9-y^{2}}}{3}\text{, }|y|\leq 3$
$y=\frac{3\sqrt{25-x^{2}}}{5}$
$y=-\frac{3\sqrt{25-x^{2}}}{5}\text{, }|x|\leq 5$