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.

$\frac{x^2}{9}+\frac{y^2}{25}=1$

Short answer

• The vertices are $(0,-5)and(0,5)$
• The foci of are $(0,-4)and(0,4)$
• 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

$\frac{x^2}{9}+\frac{y^2}{25}=1$

First we, find the greater denominator. We can see that the grater denominator, 25, is located under the y-squared, therefore the ellipse has its major axis parallel to the y-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 25 therefore it has to be that

localid="1647104958575" $$\begin{gathered} a^2=25\Rightarrow a=5 \\ and \\ {b}^2=9\Rightarrow b=3 \end{gathered}$$

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

localid="1647104995851" $(0,-a)and(0,a)$

And since $a=5$the vertices of the given ellipse arelocalid="1647105024954" $(0,-5)and(0,5)$

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.

$$\begin{gathered} c^2=a^2-b^2 \\ {c}^2=5^2-2^2 \\ {c}^2=25-4 \\ {c}^2=21 \\ c=\sqrt{21}\approx 4.58 \end{gathered}$$

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

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

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

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 x-intercepts (given by b)

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

or when we plug in $b=3$we have $(-3,0)and(3,0)$

Now draw a cure representing the ellipse through the vertices and the x-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:

localid="1647107450504" $$\begin{gathered} \frac{x^2}{9}+\frac{y^2}{25}=1 \\ \frac{y^2}{25}=1-\frac{x^2}{9} \\ {y}^2=25-\frac{25x^2}{9} \end{gathered}$$

Finally we get the two equations we have to graph:

localid="1647107457687" $y=-\sqrt{25-\frac{25x^2}{9}}andy=\sqrt{25-\frac{25x^2}{9}}$