Two Useful Transforms:
Reflection About the Line $\,y = x\,$ and
Counterclockwise Rotation by $\,90^\circ$


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 $\,y = x\,$

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:

Why does it work this way?
What's going on here?

Here's the correct thought process:

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!

Master the ideas from this section
by practicing the exercise at the bottom of this page.

When you're done practicing, move on to:
equations of ellipses in standard form: foci on the $\,y$-axis

On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.
1 2 3

(MAX is 3; there are 3 different problem types.)
Want textboxes to type in your answers? Check here: