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Q13.

Expert-verifiedFound in: Page 797

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: $(\pm 7,0)$, $(\pm 2\sqrt{6},0)$ 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:

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

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

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

This gives us

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

So vertices of the ellipse are $(\pm a,0)\u200a\u200aor\u200a\u200a\u200a(\pm 7,0)$

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 $(\pm 2\sqrt{6},0)$

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

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

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*

$\begin{array}{l}=2\left(7\right)\\ \text{\hspace{1em}}=14\end{array}$

Length of Minor axis=*2b*

$\begin{array}{l}\text{\hspace{1em}}=2\left(5\right)\\ \text{\hspace{1em}}=10\end{array}$

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

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

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

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 $(\pm c,0)$ are $(\pm 2\sqrt{6},0)$, vertices $(\pm a,0)$ are $(\pm 7,0)$.

The graph will be

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