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.

$x^2+y^2=4$

Short answer

• The vertices are on the x-axis are (2,0) and (-2,0)
  or on the y-axis (0,2) and (0,-2)
• The foci lie in the center of the circle is origin (0,0).
• 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

$x^2+y^2=4$

which written in standard form, by dividing with 14, is

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

First we, find the greater denominator. In this case the denominators are the same, 4, This means that the ellipse doesn't have a standard major axis like in the previous exercises.

An ellipse given with this equation therefore isn't longer in any direction. Therefore, the ellipse is actually circle.

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 x-intercept and y-intercept : 

Since the graph of the given equation is a circle, we don't know if the vertices are on the x-axis or on the y-axis.

Therefore, find the x-intercepts put y=0 in the equation:

$$\begin{gathered} \frac{x^2}{4}+\frac{y^2}{4}=1 \\ \frac{x^2}{4}+\frac{0^2}{4}=1 \\ \frac{x^2}{4}=1 \\ {x}^2=4 \\ x=\pm 2 \end{gathered}$$

Result is (2,0) and (-2,0)

To find the y-intercepts put x=0 in the equation:

$$\begin{gathered} \frac{x^2}{4}+\frac{y^2}{4}=1 \\ \frac{0^2}{4}+\frac{y^2}{4}=1 \\ \frac{y^2}{4}=1 \\ {y}^2=4 \\ y=\pm 2 \end{gathered}$$

Result is (0,2) and (0,-2)

Graph the Intercepts and center:

Step 3. Finding the Foci:  

To find the foci, find the intercept of the circle with the radius of a=2 around any of to x or y-intercepts found in the previous steps.

Since in the ellipse given with $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1anda>b$, the circle with radius of a drawn from the intercept that is not on the major axis gives an intercept with the major axis that is closer to the center that the vertices we can use this method to find the foci.

For example, the foci is determined by c.

In this case through a will exactly be the length of the leg connecting any of the intercepts with the center.

Therefore, the foci are in the same point and they lie in the center of the ellipse or better to say circle.

Draw a circle connecting the Intercepts:

Step 4. 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:

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

Finally, we get the two equations we have to graph:

$Y_1=-\sqrt{4-x^2}and{Y}_2=\sqrt{4-x^2}$