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Expert-verified Found in: Page 797 ### Precalculus Mathematics for Calculus

Book edition 7th Edition
Author(s) James Stewart, Lothar Redlin, Saleem Watson
Pages 948 pages
ISBN 9781337067508 # Graphing Ellipses An equation of an ellipse is given. (a) Find the vertices, foci, and eccentricity of the ellipse.(b) Determine the lengths of the major and minor axes. (c) Sketch a graph of the ellipse.$4{x}^{2}+{y}^{2}=16$.

(a).The vertices, foci and eccentricity of the ellipse $4{x}^{2}+{y}^{2}=16$ are: $\left(0,±4\right),\left(±2,0\right)$, $\left(0,±2\sqrt{5\right)}$ and $\frac{\sqrt{5}}{2}$ respectively.

(b). The length of the major axis is 8 and the length of the minor axis is 4 for the ellipse $4{x}^{2}+{y}^{2}=16$.

(c). Graph of the ellipse $4{x}^{2}+{y}^{2}=16$ is as follows: See the step by step solution

## a.Step 1. Given information.

The equation of the ellipse, $4{x}^{2}+{y}^{2}=16$

## Step 2. The vertices, foci, and eccentricity of the ellipse.

Converting the equation to standard form of ellipse $\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1$

We get,

$\begin{array}{l}\frac{4{x}^{2}}{16}+\frac{{y}^{2}}{16}=1\\ \frac{{x}^{2}}{4}+\frac{{y}^{2}}{16}=1\\ \frac{{x}^{2}}{{2}^{2}}+\frac{{y}^{2}}{{4}^{2}}=1\end{array}$

Comparing it with the standard form we get,

$\begin{array}{l}a=2\\ b=4\end{array}$

Since,a>b , the given ellipse is vertical ellipse and its focus lies on y- axis.

The coordinates of vertices are $\left(0,±b\right),\left(±a,0\right)$

Thus we get, coordinates of vertices as, $\left(0,±4\right),\left(±2,0\right)$

To find the foci, we will use the formula ${c}^{2}={a}^{2}+{b}^{2}$

By substituting the values of a,b.

We get,

$\begin{array}{c}{c}^{2}={2}^{2}+{4}^{2}\\ {c}^{2}=20\\ c=±2\sqrt{5}\end{array}$

The coordinates of foci are

$\left(0,±c\right)=\left(0,±2\sqrt{5}\right)$

The eccentricity of ellipse,

$\begin{array}{l}e=\frac{2\sqrt{5}}{4}\\ e=\frac{\sqrt{5}}{2}\end{array}$

## b.Step 1. Given information.

The equation of the ellipse, $4{x}^{2}+{y}^{2}=16$

## Step 2. The lengths of major and minor axes.

The equation of the given ellipse is $4{x}^{2}+{y}^{2}=16$

Now converting it into standard form of ellipse $\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1$

We will get,

$\begin{array}{l}\frac{4{x}^{2}}{16}+\frac{{y}^{2}}{16}=1\\ \begin{array}{l}\frac{{x}^{2}}{4}+\frac{{y}^{2}}{16}=1\\ \frac{{x}^{2}}{{2}^{2}}+\frac{{y}^{2}}{{4}^{2}}=1\end{array}\end{array}$

We will now compare it with the standard form,

Thus we get,a=2

Since a<b,

The length of the major axis: =2a

=8

Length of the minor axis:= 2b

= 4

## c.Step 1. Given information.

The equation of the given ellipse is ${x}^{2}+4{y}^{2}=16$.

## Step 2. We have to solve the given equation for y.

By converting it into the standard form of an ellipse

$\frac{{\left(x-h\right)}^{2}}{{a}^{2}}+\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1$

We will get,

$\begin{array}{l}4{x}^{2}+{y}^{2}=16\\ \begin{array}{l}⇒\frac{4{x}^{2}}{16}+\frac{{y}^{2}}{16}=1\\ ⇒\frac{{x}^{2}}{4}+\frac{{y}^{2}}{16}=1\\ ⇒\frac{{\left(x-0\right)}^{2}}{{2}^{2}}+\frac{{\left(y-0\right)}^{2}}{{4}^{2}}=1\end{array}\end{array}$

On comparing the above equation with the standard equation $\frac{{\left(x-h\right)}^{2}}{{a}^{2}}+\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1$ we get

The center(h,k) is(0,0) .

$\begin{array}{l}a=4\\ b=2\end{array}$

We will count 4 units above and below the center and 2 units to the left and right of the center.

The points are $\left(0,4\right),\left(0,-4\right),\left(2,0\right)$ and $\text{(-2,0)}$.

The graph of the ellipse will be,  ### Want to see more solutions like these? 