Getting an Ellipse by Stretching/Shrinking a Circle

 

This lesson is optional in the Precalculus course.

This lesson assumes knowledge of equations of ellipses in standard form: foci on the $\,x$-axis.

You may find yourself thinking that an ellipse looks like a ‘deformed circle’:
you know, just grab the ‘ends’ and stretch or compress the circle.
And—you'd be correct, as this lesson shows!

By using earlier work on horizontal and vertical stretching/shrinking of graphs,
we get the equation $\displaystyle\,\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\,$ easily, as follows:
  • Start with the circle of radius $\,1\,,$ centered at the origin, which has equation $\,x^2 + y^2 = 1\,.$
  • Replace every $\,x\,$ in the circle equation by $\displaystyle\,\frac xa\,$:
    $$ \begin{gather} \left(\frac xa\right)^2 + y^2 = 1\cr \frac{x^2}{a^2} + y^2 = 1 \end{gather} $$ This moves a point $\,(x,y)\,$ on the circle to a new point $\,(ax,y)\,$ on an ellipse.
    For example, the point $\,(1,0)\,$ on the circle moves to $\,(a,0)\,$ on the ellipse.
    If $\,a > 1\,,$ it is a horizontal stretch.
    If $\,0 < a < 1\,,$ it is a horizontal shrink.
  • Similarly, replace every $\,y\,$ with $\displaystyle\,\frac yb\,$ in the equation $\,x^2 + y^2 = 1\,,$ giving the new equation $\displaystyle\,x^2 + \frac{y^2}{b^2} = 1\,.$
    This moves a point $\,(x,y)\,$ on the circle to a new point $\,(x,by)\,$ on an ellipse.
    For example, the point $\,(0,1)\,$ on the circle moves to $\,(0,b)\,$ on the ellipse.
    If $\,b > 1\,,$ it is a vertical stretch.
    If $\,0 < b < 1\,,$ it is a vertical shrink.
  • By replacing both $\,x\,$ by $\displaystyle\,\frac xa\,$ and $\,y\,$ by $\displaystyle\,\frac yb\,,$ we get the equation $\,\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\,.$
circle of radius $\,1\,,$ in blue:
$x^2 + y^2 = 1$

ellipse, in green:
$$\color{green}{\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1}$$ $\,a > 1\,,$ horizontal stretch
$\,0 < b < 1\,,$ vertical shrink
circle of radius $\,1\,,$ in blue:
$x^2 + y^2 = 1$

horizontal stretch, in red:
$\displaystyle\frac{x^2}{a^2} + y^2 = 1\,,$ for $\,a > 1$

horizontal shrink, in green:
$\displaystyle\frac{x^2}{a^2} + y^2 = 1\,,$ for $\,0 < a < 1$




circle of radius $\,1\,,$ in blue:
$x^2 + y^2 = 1$

vertical stretch, in red:
$\displaystyle x^2 + \frac{y^2}{b^2} = 1\,,$ for $\,b > 1$

vertical shrink, in green:
$\displaystyle x^2 + \frac{y^2}{b^2} = 1\,,$ for $\,0 < b < 1$

Transformations involving $\,y\,$
for Equations that ARE of the Form $\,y = f(x)\,$

People who have thoroughly studied horizontal and vertical stretching/shrinking
may be a bit perplexed by the treatment of the transformations involving $\,y\,$ in this section.
If this is you, then keep reading!

In that earlier section, we were dealing with a very specific type of equation involving $\,x\,$ and $\,y\,$: an equation of the form $\,y = f(x)\,.$
Such an equation has $\,y\,$ all by itself on one side; the other side involves only $\,x\,.$
For equations of the form $\,y = f(x)\,,$ it was easier to treat transformations involving $\,x\,$ and $\,y\,$ differently.
We'll quickly review those differences next.

Transformations involving $y$ for equations of the form $\,y = f(x)\,$:

Transformations involving $x$ for equations of the form $\,y = f(x)\,$:

Transformations involving $\,y\,$
for Equations that ARE NOT of the Form $\,y = f(x)\,$

Sometimes you need to apply graphical transformations to equations that can't be put in the form $\,y = f(x)\,.$
If the graph of an equation doesn't pass a vertical line test, then it can't be put in the form $\,y = f(x)\,.$

The ellipse equation $\displaystyle\,\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\,$ is such an equation.
Its graph doesn't pass the vertical line test.
Here, it is possible to solve for $\,y\,,$ but it introduces an annoying square root and ‘plus or minus’ sign.

To apply graphical transformations to any equation in $\,x\,$ and $\,y\,,$
(no matter how mixed-up the variables might be),
just use the ‘replace every...’ language for both $\,x\,$ and $\,y\,.$
(Then, both cases are equally counter-intuitive!)

For example:
If we want every $y$-value in a graph to be multiplied by $\,3\,,$
then replace every $\,y\,$ in the original equation by $\,y\,$ divided by $\,3\,.$

This more general technique certainly works in the ‘$\,y = f(x)\,$’ case, but isn't needed:

EXAMPLE:
Start with the graph of $\,y = x^2\,.$
Multiply each $y$-value of points on the graph by $\,3\,$ (a vertical stretch).
What is the equation of the resulting graph?

ANSWER:
$\displaystyle\,\frac{y}{3} = x^2\,$
Replace every $\,y\,$ by $\,\frac{y}{3}\,.$
Of course, this gives the same result—just multiply both sides by $\,3\,$ to get $\,y = 3x^2\,$!

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

When you're done practicing, move on to:
(optional) two useful transforms:
reflection about the line $\,y = x\,$ and
counterclockwise rotation by $\,90^\circ$


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