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Q18.

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.**

$4{x}^{2}+{y}^{2}=16$.

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

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

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 $(0,\pm b),(\pm a,0)$

Thus we get, coordinates of vertices as, $(0,\pm 4),(\pm 2,0)$

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

By substituting the values of *a,b.*

We get,

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

The coordinates of foci are

$(0,\pm c)=(0,\pm 2\sqrt{5})$

The eccentricity of ellipse,

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

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

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

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

By converting it into the standard form of an ellipse

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

We will get,

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

On comparing the above equation with the standard equation $\frac{{(x-h)}^{2}}{{a}^{2}}+\frac{{(y-k)}^{2}}{{b}^{2}}=1$ 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 $(0,4),(0,-4),(2,0)$ and $\text{(-2,0)}$.

The graph of the ellipse will be,

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