Example 4 from Algebra and Trigonometry, 12.1 The Ellipse

Graphing an Ellipse Centered at the Origin from an Equation Not in Standard Form

Graph the ellipse given by the equation $4 x^{2} + 25 y^{2} = 100.$ Rewrite the equation in standard form. Then identify and label the center, vertices, co-vertices, and foci.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

1
Solution
First, use algebra to rewrite the equation in standard form.

$\begin{array}{l} 4 x^{2} + 25 y^{2} = 100 \\ \frac{4 x^{2}}{100} + \frac{25 y^{2}}{100} = \frac{100}{100} \\ \frac{x^{2}}{25} + \frac{y^{2}}{4} = 1 \end{array}$
Next, we determine the position of the major axis. Because $25 > 4,$ the major axis is on the x-axis. Therefore, the equation is in the form $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1,$ where $a^{2} = 25$ and $b^{2} = 4.$ It follows that:
- the center of the ellipse is $(0, 0)$
- the coordinates of the vertices are $(\pm a, 0) = (\pm \sqrt{25}, 0) = (\pm 5, 0)$
- the coordinates of the co-vertices are $(0, \pm b) = (0, \pm \sqrt{4}) = (0, \pm 2)$
- the coordinates of the foci are $(\pm c, 0),$ where $c^{2} = a^{2} - b^{2} .$ Solving for $c,$ we have:
$\begin{array}{l} c = \pm \sqrt{a^{2} - b^{2}} \\ = \pm \sqrt{25 - 4} \\ = \pm \sqrt{21} \end{array}$

Therefore the coordinates of the foci are $(\pm \sqrt{21}, 0) .$

Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse.

Figure 9

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Graph the ellipse given by the equation $49 x^{2} + 16 y^{2} = 784.$ Rewrite the equation in standard form. Then identify and label the center, vertices, co-vertices, and foci.

How did it go?

Solution

Your turn

Solution

Your turn