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.
Reflection About the Line $\,y = x\,$

As discussed in an earlier section, reflection about the line $\,y = x\,$
switches the coordinates of a point:
$\,(x,y)\,$ moves to $\,(y,x)\,$.

For example, suppose that reflection about $\,y = x\,$ is applied to the ellipse
$\displaystyle\,\frac{x^2}{a^2} +\frac{y^2}{b^2} = 1\,$,
which has its foci on the $x$axis.
Then
 $\,(c,0)\,$ moves to $\,(0,c)\,$
 $\,(c,0)\,$ moves to $\,(0,c)\,$
So, the reflected ellipse has its foci on the $y$axis.

In an equation, reflection about the line $\,y = x\,$ is accomplished by switching the variables $\,x\,$ and $\,y\,$.

For example, switching $\,x\,$ and $\,y\,$ turns
$$\frac{x^2}{a^2} +\frac{y^2}{b^2} = 1$$
into:
$$\frac{y^2}{a^2} +\frac{x^2}{b^2} = 1$$
As you'll see in the next section, this new equation is precisely
the ellipse with center at the origin and
foci along the $y$axis!
So easy!
Counterclockwise Rotation by $\,90^\circ$
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!


Here are the correct replacements in an equation,
to rotate its graph counterclockwise by $\,90^\circ\,$:
 replace every $\,x\,$ by $\,y\,$
 replace every $\,y\,$ by $\,x\,$
For example, applying a $\,90^\circ\,$ counterclockwise rotation to the equation $\,y = \sqrt{x}\,$ gives
the following:
$$
\begin{gather}
\overbrace{y}^{\text{replace by $\,x\,$}} = \sqrt{\overbrace{x}^{\text{replace by $\,y\,$}}}\cr\cr
x = \sqrt{\vphantom{h}\, y\, }
\end{gather}
$$
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!


So, what's going on here?
In a graph, we want $\,(x,y)\,$ to move to $\,(y,x)\,$.
In the equation, this is accomplished by:
 replacing every $\,x\,$ by $\,y\,$
 replacing every $\,y\,$ by $\,x\,$
Why does it work this way?
What's going on here?
Here's the correct thought process:

Every equation in two variables ($\,x\,$ and $\,y\,$)
can be written in the form $\,f(x,y) = 0\,$.
For example, the equation $\,y = x^2\,$ can be written as $\,y  x^2 = 0\,$,
so defining $\,f(x,y) := y  x^2\,$ does the job.

The graph of the original equation, $\,f(x,y) = 0\,$, is (by definition)
$$
\{\ (x,y) \ \ \ \ f(x,y) = 0\ \} \qquad (*)
$$
Read (*) as: the set of all points $\,(x,y)\,$ with the property that $\,f(x,y) = 0\,$.

The set of points we want, after
rotating the points in (*) counterclockwise by $\,90^\circ\,$, is:
$$
\{\ (y,x) \ \ \ \ f(x,y) = 0\ \} \qquad (**)
$$

Define $\,\hat x := y\,$ and $\,\hat y := x\,$, to view the points in (**) in a more natural way:
$$
\{\ (\ \overbrace{\strut y}^{\hat x}\ ,\ \overbrace{\strut x}^{\hat y}\ ) \ \ \ \ f(x,y) = 0\ \} \qquad (**)
$$
(Read $\,\hat x\,$ as ‘$x$ hat’ and $\,\hat y\,$ as ‘$y$ hat’.)

Solve for $\,x\,$ and $\,y\,$ in terms of $\,\hat x\,$ and $\,\hat y\,$:
 $\,x = \hat y\,$
 $\,y = \hat x\,$

Rename (**):
$$
\begin{align}
\{(\overbrace{\strut y}^{\hat x}&,\overbrace{\strut x}^{\hat y}) \ \ \ \ f(x,y) = 0\}\cr\cr
&= \{(\hat x,\hat y)\ \ \ \ f(\hat y,\hat x) = 0\}\cr\cr
&= \{(x,y)\ \ \ \ f(y,x) = 0\}\qquad \qquad \text{(return to the dummy variables $\,x\,$ and $\,y\,$)}
\end{align}
$$

How do we get from the original equation ‘$\,f(x,y) = 0\,$’ to
the new (desired) equation ‘$\,f(y,x) = 0\,$’?
Answer: Replace every $\,x\,$ by $\,y\,$, and replace every $\,y\,$ by $\,x\,$!
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!