Answer
$y=\dfrac{b\sqrt{a^2-x^2}}{a}$
Work Step by Step
Need to solve for $y$.
We have $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$
This can be written in the form of $y$ as: $\dfrac{y^2}{b^2}=1-\dfrac{x^2}{a^2}$
or, $\dfrac{y}{b}=\sqrt{1-\dfrac{x^2}{a^2}}$
or, $y=b \sqrt{\dfrac{{a^2-x^2}}{a^2}}$
Hence, $y=\dfrac{b\sqrt{a^2-x^2}}{a}$
