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

$x^2+\frac{y^2}{16}=1$

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

$x^2+\frac{y^2}{16}=1$

First we, find the greater denominator. We can see that the grater denominator, 16, 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).

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="1647105678988" $$\begin{gathered} a^2=16\Rightarrow a=4 \\ and \\ {b}^2=1\Rightarrow b=1 \end{gathered}$$

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

localid="1647105694756" $(0,-a)and(0,a)$

And since the vertices of the given ellipse arelocalid="1647105715323" $(0,-4)and(0,4)$

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.

$$\begin{gathered} c^2=a^2-b^2 \\ {c}^2=4^2-1^2 \\ {c}^2=16-1 \\ {c}^2=15 \\ c=\sqrt{15}\approx 3.87 \end{gathered}$$

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

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

Therefore they are$(0,-\sqrt{15})and(0,\sqrt{15})$

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 $b=1$we have $(-1,0)and(1,0)$

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:

$$\begin{gathered} \frac{x^2}{25}+\frac{y^2}{4}=1 \\ \frac{y^2}{16}=1-x^2 \\ {y}^2=16-16x^2 \end{gathered}$$

Finally we get the two equations we have to graph:

$y=-\sqrt{16-16x^2}andy=\sqrt{16-16x^2}$