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

$\frac{x^2}{25}+\frac{y^2}{4}=1$

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

$\frac{x^2}{25}+\frac{y^2}{4}=1$

First we, find the greater denominator. We can see that the grater denominator, 25, 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).

Any ellipse equation has the following form:

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

but in both cases $a>b$.

Since the greater denominator is 25 therefore it has to be that

$$\begin{gathered} a^2=25\Rightarrow a=5 \\ and \\ {b}^2=4\Rightarrow b=2 \end{gathered}$$

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

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

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

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:

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

Therefore they are$(-\sqrt{21},0)and(\sqrt{21},0)$

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 $b=2$we have $(0,-2)and(0,2)$

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

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:

$$\begin{gathered} \frac{x^2}{25}+\frac{y^2}{4}=1 \\ \frac{y^2}{4}=1-\frac{x^2}{25} \\ {y}^2=4-\frac{4x^2}{25} \end{gathered}$$

Finally we get the two equations we have to graph:

$y=-\sqrt{4-\frac{4x^2}{25}}andy=\sqrt{4-\frac{4x^2}{25}}$