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Q13.

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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. $\frac{{x}^{2}}{49}+\frac{{y}^{2}}{25}=1$

(a).The vertices, foci and eccentricity of the ellipse $\frac{{x}^{2}}{49}+\frac{{y}^{2}}{25}=1$ are: $\left(±7,0\right)$, $\left(±2\sqrt{6},0\right)$ and $\frac{2\sqrt{6}}{7}$ respectively.

(b). The length of the major axis is 14 and the length of the minor axis is 10 for the ellipse $\frac{{x}^{2}}{49}+\frac{{y}^{2}}{25}=1$.

(c). Graph of the ellipse $\frac{{x}^{2}}{49}+\frac{{y}^{2}}{25}=1$ is as follows:

See the step by step solution

## a.Step 1. Given information.

The equation of the ellipse, $\frac{{x}^{2}}{49}+\frac{{y}^{2}}{25}=1$

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

If we compare the given equation $\frac{{x}^{2}}{49}+\frac{{y}^{2}}{25}=1$ with standard form

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

This gives us

$\begin{array}{l}h=0,\\ k=0\\ {a}^{2}=49,\\ ⇒a=7\\ {b}^{2}=25.\\ ⇒b=5\end{array}$

So vertices of the ellipse are $\left(±a,0\right) or \left(±7,0\right)$

Foci are $(\pm c,0)$,where ${c}^{2}={a}^{2}-{b}^{2}$

$c=\sqrt{{a}^{2}-{b}^{2}}$

By substituting values we will get ${c}^{2}=49-25$

Therefore Foci are $\left(±2\sqrt{6},0\right)$

Eccentricity $e=\frac{c}{a}.$ We have values of both, by substituting themwe get $e=\frac{2\sqrt{6}}{7}$

## b.Step 1. Given information.

The equation of the ellipse, $\frac{{x}^{2}}{49}+\frac{{y}^{2}}{25}=1$

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

The equation of the given ellipse is $\frac{{x}^{2}}{49}+\frac{{y}^{2}}{25}=1$

By comparing the given equation with the general form we found that

$\begin{array}{l}a=7\\ b=5\end{array}$

Length of Major axis =2a

$\begin{array}{l}=2\left(7\right)\\ \text{ }=14\end{array}$

Length of Minor axis=2b

$\begin{array}{l}\text{ }=2\left(5\right)\\ \text{ }=10\end{array}$

## c.Step 1. Given information.

The equation of the given ellipse is $\frac{{x}^{2}}{49}+\frac{{y}^{2}}{25}=1$.

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

From the equation of ellipse, we can see that denominator of ${x}^{2}$ is greater than that of ${y}^{2}$. So this Would be a Horizontal ellipse. By comparing the given equation with the standard form equation $\frac{{\left(x-h\right)}^{2}}{{a}^{2}}+\frac{{\left(y-k\right)}^{2}}{{b}^{2}}=1$

$\frac{{x}^{2}}{49}+\frac{{y}^{2}}{25}=1$

We will get:

$\begin{array}{l}h=0\\ k=0\\ {a}^{2}=49\\ a=7\\ {b}^{2}=25\\ b=5\\ c=\sqrt{{a}^{2}-{b}^{2}}\\ c=\sqrt{24}\\ c=2\sqrt{6}\end{array}$

Foci of the ellipse $\left(±c,0\right)$ are $\left(±2\sqrt{6},0\right)$, vertices $\left(±a,0\right)$ are $\left(±7,0\right)$.

The graph will be