# How do you graph the ellipse #(x+5)^2/16+(y-4)^2/1=1# and find the center, the major and minor axis, vertices, foci and eccentricity?

##### 1 Answer

Please see below.

#### Explanation:

whose center is

vertices are

Eccentricity

and focii are at a distance of

Hence, in

center is

Vertices are

Eccentricity is

and focii are

They appear as follows:

graph{((x+5)^2+16(y-4)^2-16)((x+5)^2+(y-4)^2-0.01)((x+5)^2+(y-5)^2-0.01)((x+5)^2+(y-3)^2-0.01)((x+1)^2+(y-4)^2-0.01)((x+9)^2+(y-4)^2-0.01)((x+5+sqrt3)^2+(y-4)^2-0.01)((x+5-sqrt3)^2+(y-4)^2-0.01)=0 [-10.12, -0.12, 1.46, 6.46]}