Question

In Problems 17–26, find the vertices and foci of each ellipse. Graph each equation by hand. Verify your graph using a graphing utility.

$x^2+9y^2=18$

Short answer

• The vertices are$(-3\sqrt{2},0)and(3\sqrt{2},0)$
• The foci of are$(-4,0)and(4,0)$
• Graph is

Step-by-step solution

Step 1. Given information and Analyze the Equation:  

We want to find the vertices and foci of the ellipse given with its equation, then graph by hand and check that graph is correct using a graphing utility.

The equation we have to analyze in order to find the vertex and foci is

$x^2+9y^2=18$

By dividing with 18 we get the standard form:

$\frac{x^2}{18}+\frac{y^2}{2}=1$

First we, find the greater denominator. We can see that the grater denominator, 18, is located under the x-squared, therefore the ellipse has its major axis parallel to the x-axis.

Since the term squared have no numbers squared, only the variables x or y. the center of the ellipse is located in the origin (0,0).

Step 2. Finding the vertices:  

Any ellipse equation has the following form:

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

but in both cases $a>b$

Since the greater denominator is 18 therefore it has to be that

$$\begin{gathered} a^2=18\Rightarrow a=3\sqrt{2}\approx 4.24 \\ and \\ {b}^2=2\Rightarrow b=\sqrt{2}\approx 1.41 \end{gathered}$$

An ellipse with its major axis parallel to the x-axis has vertices in the points

$(-a,0)and(a,0)$

And since $a=3\sqrt{2}$the vertices of the given ellipse are

$(-3\sqrt{2},0)and(3\sqrt{2},0)$

Step 3. Finding the Foci:  

The distance from the origin of the ellipse to the foci is c.

Looking at the following picture:

The triangle ABC is right triangle therefore the Pythagoras theorem holds and is used to find c when we have a and b.

$$\begin{gathered} c^2=a^2-b^2 \\ {c}^2={\left(3\sqrt{2}\right)}^2-{\left(\sqrt{2}\right)}^2 \\ {c}^2=18-2 \\ {c}^2=16 \\ c=4 \end{gathered}$$

The foci are also located on the major axis and have the following coordinates:

$(-c,0)and(c,0)$

Therefore they are$(-4,0)and(4,0)$

Step 4. Graphing the equation by hand: 

To graph the equation by hand first graph the found points: vertices, center of the ellipse and its y-intercepts (given by b)

The y-intercepts are$(0,-b)and(0,b)$

or when we plug in $b=\sqrt{2}$we have$(0,-\sqrt{2})and(0,\sqrt{2})$

Now draw a cure representing the ellipse through the vertices and the y-intercepts and add the foci:

Step 5. Using a graphing utility: 

Since the graph doesn't pass the vertical line test, the equation of the ellipse is not a function.

In order to graph the ellipse we will need to graph two functions obtaining by first expressing $y^2$ on the side and then taking square root to get two equations:

$$\begin{gathered} \frac{x^2}{18}+\frac{y^2}{2}=1 \\ \frac{y^2}{2}=1-\frac{x^2}{18} \\ {y}^2=2-\frac{x^2}{9} \end{gathered}$$

Finally we get the two equations we have to graph:

$y=-\sqrt{2-\frac{x^2}{9}}andy=\sqrt{2-\frac{x^2}{9}}$