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.

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

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

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

We get,

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

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

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

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

We will get,

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

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

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,

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

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) .

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,