In Problems 27–38, find an equation for each ellipse. Graph the equation by hand.

Center at (0, 0); focus at (0, 1); vertex at (0, -2)

• Center at (0, 0)
• Focus at (0, 1)
• Vertex at (0, -2)

We want to use the given Information to find the equation of the ellipse with these property. Then graph the ellipse by hand.

Since the ellipse has its center at the origin (0,0) the major axis must lie either on the y-axis or on the x-axis.

On the major axis lie the foci and vertices of the given ellipse.

Now by noticing that the center, focus and vertex both have the same x-coordinates and that coordinate is zero, and since they have to lie on the major axis, we can conclude that the major axis is equal to the y-axis.

We can also make a drawing to visualize:

The standard form for an ellipse equation, with its major axis parallel to the x-axis is$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1witha>b$

Therefore, in order to find the equation of the ellipse we need to find a and b .

An ellipse with its major axis being the x-axis has the following vertices:

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

and the following foci:

$$\begin{gathered} (0,-c)and(0,c) \\ where{c}^2=a^2-b^2 \end{gathered}$$

Since one vertex is given: $(0,-2)$it follows that role="math" localid="1647231559243" $a=2$.

Since the term needed in the equation is a² we finally get

role="math" localid="1647231612231" $a^2=2^2=4$

Similarly, since the focus (0,1) is given, it follows that$c=3$

To find b² use the formula mentioned in the step:

$$\begin{gathered} c^2=a^2-b^2 \\ {b}^2=a^2-c^2 \\ {b}^2=2^2-1^2 \\ {b}^2=4-1 \\ {b}^2=3 \\ b=\pm \sqrt{3} \end{gathered}$$

Finally, a²=4 and b²=3gives the equation of the ellipse.

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

Plug in the values to get$\frac{x^2}{3}+\frac{y^2}{4}=1$

To graph the ellipse remember that from the given focus and vertex, we can obtain the other focus and vertex by changing the sign of the y-coordinates.

We can do this because the ellipse is symmetrical with respect to the x-axis.

Therefore the second vertex is $(0,2)$.

Similarity, since the given ellipse is symmetric with respect to the y-axis (its centre in the origin) the x-intercepts are determine by b and they are:

$$\begin{gathered} (-b,0)and(b,0) \\ or \\ (-\sqrt{3},0)and(\sqrt{3},0) \end{gathered}$$

Therefore, by graphing all the vertices and x-intercepts and connecting them we get: