Chapter 10.3, Problem 21AYU

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

$4x^2+y^2=16$

To determine

(a & b)

To find: The vertices and foci of the given equation of the ellipse.

Answer to Problem 21AYU

Solution:

Vertices: $(0,4)\text{and}(0,-4)$ and Foci: $\left(0,2\surd 3\right)\text{and}\left(0,-2\surd 3\right)$.

Explanation of Solution

Given:

$4x²+y²=16$

Formula used:

Equation	Center	Major axis	Foci	Vertices
$\begin{array}{l}\frac{x^2}{b^2}+\frac{y^2}{c^2}=1\\ a>b>0\text{and}b²=a²-c²\end{array}$	$\left(0,0\right)$	$y\text{-axis}$	$\left(0,c\right)$	$\left(0,a\right)$
	$\left(0,0\right)$	$y\text{-axis}$	$\left(0,-c\right)$	$\left(0,-a\right)$

Calculation:

Dividing the given equation $4x²+y²=16$ by 16, we get;

$\frac{x^2}{4}+\frac{y^2}{16}=1$

The center of the ellipse is at origin $\left(0,0\right)$;

$a=\sqrt{16}=4;b=\sqrt{4}=2$

The vertices are $\left(0,\pm a\right)$, hence vertices: $\left(0,4\right)\text{and}\left(0,-4\right)$.

To find the value of $c$,

$c²=4²-2²=16-4=16;c=2\surd 3$

Hence the foci are $\left(0,2\surd 3\right)\text{and}\left(0,-2\surd 3\right)$.