Chapter 11, Problem 2T

To determine

To find: The vertices, foci and the length of the major and minor axis of the ellipse $\frac{x^2}{16}+\frac{y^2}{4}=1$ and sketch the graph.

Answer to Problem 2T

The focus of the ellipse $\frac{x^2}{16}+\frac{y^2}{4}=1$ are $F\left(\pm 2\sqrt{3},0\right)$.

The vertices of the ellipse $\frac{x^2}{16}+\frac{y^2}{4}=1$ are $\left(\pm 4,0\right)$.

The length of major axis in the ellipse $\frac{x^2}{16}+\frac{y^2}{4}=1$ is 8.

The length of minor axis in the ellipse $\frac{x^2}{16}+\frac{y^2}{4}=1$ is 4.

Explanation of Solution

Definition used:

“The equation of the ellipse center at the origin with vertices $(\pm a,0)$ and foci $F(\pm c,0)$ is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where $c=\sqrt{a^2-b^2}$, $a>b>0$ and the length of major and minor axes of the ellipse $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$ are $2a$ and $2b$ respectively”.

The given equation is in standard form of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.

Therefore, compare the equation Use online graphing calculator and obtain the graph of $x^2=-12y$ as shown below in Figure 1.

 with the equation of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.

So , clearly the value of $a^2=16$ and $b^2=4$.

Compute the value of $a$.

$\begin{array}{c}{a}^2=16\\ a=\sqrt{16}\\ a=\pm 4\end{array}$

Therefore, the value of $a$ is $\pm 4$.

Compute the value of $b$.

$\begin{array}{c}{b}^2=4\\ b=\sqrt{4}\\ b=\pm 2\end{array}$

Therefore, the value of $b$ is $\pm 2$.

Compute the value of $c$.

$\begin{array}{c}c=\sqrt{a^2-b^2}\\ =\sqrt{16-4}\\ =\sqrt{12}\\ =\pm 2\sqrt{3}\end{array}$

Therefore, the value of $c$ is $\pm 2\sqrt{3}$.

From by the definition stated above, the focus and vertices of the ellipse are $F(\pm c,0)$ and $(\pm a,0)$ respectively.

Therefore, the focus and vertices of the ellipse $\frac{x^2}{16}+\frac{y^2}{4}=1$ are $F\left(\pm 2\sqrt{3},0\right)$ and $\left(\pm 4,0\right)$.

The length of major axis in the ellipse $\frac{x^2}{16}+\frac{y^2}{4}=1$ is 8.

The length of minor axis in the ellipse $\frac{x^2}{16}+\frac{y^2}{4}=1$ is 4.

Use online graphing calculator and obtain the graph of $\frac{x^2}{16}+\frac{y^2}{4}=1$ as shown below in Figure 1.

From figure 1, observed that the focus and vertices of the ellipse $\frac{x^2}{16}+\frac{y^2}{4}=1$ are $F\left(\pm 2\sqrt{3},0\right)$ and $\left(\pm 4,0\right)$.