Question 10.2.4: Graphing an ellipse centered at (h, k) Sketch the graph and ...
Graphing an ellipse centered at (h, k)
Sketch the graph and identify the foci of the ellipse
\frac{(x-3)^{2}}{25}+\frac{(y+1)^{2}}{9}=1The graph of this equation is a translation of the graph of
\frac{x^{2}}{25}+\frac{y^{2}}{9}=1three units to the right and one unit downward. The center of the ellipse is (3, -1). Since a^{2}=25, the vertices lie five units to the right and five units to the left of (3, -1). So the ellipse goes through (8, -1) and (-2, -1). Since b^{2}=9, the graph includes points three units above and three units below the center. So the ellipse goes through (3, 2) and (3, -4), as shown in Fig. 10.24. Because c^{2}=25-9=16, c=\pm 4 . The major axis is on the horizontal line y = -1. Thus the foci are found four units to the right and four units to the left of the center at (7, -1) and (-1, -1).
To check the location of the ellipse, graph
y_{1}=-1+\sqrt{9-9(x-3)^{2} / 25} and y_{2}=-1-\sqrt{9-9(x-3)^{2} / 25}
on a graphing calculator, as in Fig. 10.25.
