Chapter 10, Problem 5CT

To determine

The equation of the ellipse and graph of the equation with center $(0,0)$, vertex $(0,-4)$, and focus $(0,3)$

Answer to Problem 5CT

Solution:

The equation of the ellipse with center $(0,0)$, vertex $(0,-4)$, and focus $(0,3)$ is $\frac{x^2}{7}+\frac{y^2}{16}=1$ and the graph of the equation is

  

Explanation of Solution

Given information:

The ellipse has center $(0,0)$, vertex $(0,-4)$, and focus $(0,3)$

The center $(0,0)$, vertex $(0,-4)$, and focus $(0,3)$ all lie on the vertical line $x=0$

Thus, the major axis is $y-$ axis.

The distance from the center $(0,0)$ to a focus $(0,3)$ is $c=3$, and the distance from the center $(0,0)$ to a vertex $(0,-4)$ is $a=4$ .

  $b^2=a^2-c^2=4^2-3^2=16-9=7$ .

  $\Rightarrow b=\sqrt{7}$

The general form of equation of ellipse is $\frac{{(x-h)}^2}{b^2}+\frac{{(y-k)}^2}{a^2}=1$

Substitute $h=0,k=0,a=4$ and $b=\sqrt{7}$ ,

  $\frac{{(x-0)}^2}{{\left(\sqrt{7}\right)}^2}+\frac{{(y-0)}^2}{4^2}=1$

  $\Rightarrow \frac{x^2}{7}+\frac{y^2}{16}=1$ .

To graph the equation, use the center $(h,k)=(0,0)$ to locate the vertices. The major axis is the $y-$ axis and $a=4$, the vertices are $V_1=(0,4)$ and $V_2=(0,-4)$ .

Since $c=3$, the foci are $F_1=(0,3)$ and $F_2=(0,-3)$ . Also $b=\sqrt{7}$ then two points left and right of the center are $\left(h\pm b,k\right)$, which are $\left(-\sqrt{7},0\right)$ and $\left(\sqrt{7},0\right)$ .

By using all the above information, the graph of the ellipse $\frac{x^2}{7}+\frac{y^2}{16}=1$ is as follows: