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

This section is optional in the Precalculus course.

Once we have the ellipse equation for foci on the $x$-axis,

$$\frac{x^2}{a^2} +\frac{y^2}{b^2} = 1\,,$$

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 $\,\frac{x^2}{a^2} +\frac{y^2}{b^2} = 1\,,$ which has its foci on the $x$-axis. Then:

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 $\,90^\circ$

counterclockwise rotation by 90 degrees

As illustrated by the sketch above, 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:

$$ \begin{gather} \overbrace{y}^{\text{replace by $\,-x\,$}} = \sqrt{\overbrace{x}^{\text{replace by $\,y\,$}}}\cr\cr -x = \sqrt{\vphantom{h}\, y\, } \end{gather} $$ rotating the square root graph counterclockwise 90 degrees

Look at the plot results above 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?

The Correct Thought Process

Here's the correct thought process:

Write the Equation in the Form $\,f(x,y) = 0$

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.

Graph of the Equation $\,f(x,y) = 0$

The graph of the original equation, $\,f(x,y) = 0\,,$ is (by definition):

$$ \{\ (x,y) \ \ |\ \ f(x,y) = 0\ \} \quad (*) $$

Read ($*$) as:  the set of all points $\,(x,y)\,$ with the property that $\,f(x,y) = 0\,.$

The Set of Points We Want

The set of points we want, after rotating the points in ($*$) counterclockwise by $\,90^\circ\,,$ is:

$$ \{\ (-y,x) \ \ |\ \ f(x,y) = 0\ \} \quad (**) $$

Rewrite the Set of Points

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 $\,\hat y\,$:

$$ \begin{gather} x = \hat y\cr \text{and}\cr y = -\hat x \end{gather} $$

Rename ($**$):

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

The Conclusion

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 $\,-x\,$!

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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!

Master the ideas from this section by practicing below:

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When you're done practicing, move on to:

Equations of Ellipses in Standard Form: Foci on the $y$-axis
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Concept Practice

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