Graphing Ellipses Centered at (h,k)

Example

Graph $\frac{\left(x-2\right)^2}{9}+\frac{\left(y+1\right)^2}{16}=1$. Give the foci, domain, and range.

Solution

The graph of this equation is an ellipse centered at (2, −1). Ellipses always have a > b. For this ellipse a = 4 and b = 3. Since a = 4 is associated with $y^2$,the vertical vertices of the ellipse are on the vertical line through (2, −1). In this book, it is only the vertices on the major axis that are called vertices, but usually an ellipse is thought to have 4 vertices, not two (with the remaining two being the points on the minor axis.) 

To find the vertices, locate two points on the vertical line through (2, −1), one 4 units up from (2, −1), and one 4 units down. This gives the vertices (2, 3) and (2, −5).

Two other points on the ellipse, also sometimes considered vertices, are located a distance of b = 3 units to the left and right of the center, are (–1, –1) and (5, –1).

To determine the foci, we use the equation $c^2=a^2-b^2$.

$c^2=a^2-b^2$

$c=\sqrt[]{7}$

The foci are at $\left(2,-1+\sqrt[]{7}\right)$ and $\left(2,-1-\sqrt[]{7}\right)$.

The domain is $\left[-1,5\right]$ and the range is $\left[-5,3\right]$.

$c^2=16-9=7$

Note

As suggested by the graphs in this section, an ellipse is symmetric with respect to its major axis, its minor axis, and its center. If a = b in the equation of an ellipse, then the graph is a circle.