This section is optional in the Precalculus course.
Once we have the ellipse equation for foci on the $x$axis $\displaystyle\,\left(\frac{x^2}{a^2} +\frac{y^2}{b^2} = 1\right)\,,$
we don't have to go through that long and tedious derivation a second time to get the equation for foci on the $y$axis.
There are easier ways!
If desired, we can use appropriate transforms to get the new graph.
This section discusses two transforms that can be used.
As illustrated by the sketch at right, rotating counterclockwise by $\,90^\circ\,$ moves $\,(x,y)\,$ to $\,(y,x)\,.$ Think of the green triangle as a wedge of wood; grab it and rotate it, so the side on the $x$axis moves to the $y$axis. It ‘turns into’ the red triangle! How can an equation be ‘adjusted’ to accomplish this $\,90^\circ\,$ counterclockwise rotation? Be very careful! Since $\,(x,y)\,$ moves to $\,(y,x)\,,$ you might be tempted to think that, in the original equation, you should replace every $\,x\,$ by $\,y\,$ and every $\,y\,$ by $\,x\,.$ But, THIS IS WRONG! 

Look at the
plot results at right from WolframAlpha. The equation $\,y = \sqrt x\,$ is plotted in blue. The equation $\,x = \sqrt y\,$ is plotted in purple. Notice the beautiful $\,90^\circ\,$ counterclockwise rotation (from the blue curve to the purple curve), as desired! 

In a graph, we want $\,(x,y)\,$ to move to $\,(y,x)\,.$
In the equation, this is accomplished by:
Here's the correct thought process:
EXAMPLE
Rotate the ellipse
$\displaystyle\,\frac{x^2}{a^2} +\frac{y^2}{b^2} = 1\,$
counterclockwise by $\,90^\circ\,$:
$$
\begin{gather}
\frac{y^2}{a^2} +\frac{(x)^2}{b^2} = 1\cr\cr
\frac{y^2}{a^2} + \frac{x^2}{b^2} = 1
\end{gather}
$$
As before, you'll see in the next section that this is precisely
the ellipse with center at the origin and
foci along the $y$axis!
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
IN PROGRESS 