How do you graph the ellipse #(x+2)^2/16+(y-5)^2/20=1# and find the center, the major and minor axis, vertices, foci and eccentricity?

1 Answer
Jun 4, 2017

Answer:

Please see below.

Explanation:

The centre of the ellipse #(x-h)^2/a^2+(y-k)^2/b^2=1# is #(h,k)#.

The two axes are #2a# and #2b# and major axis is the greater of the two.

Vertices are #(h+-a,k)# and #(h,k+-b)#. Here those along major axis are called vertices and those along minor axis are called covertices.

#(x+2)^2/16+(y-5)^2/20=1# can be written as #(x+2)^2/4^2+(y-5)^2/(2sqrt5)^2=1#

hence centre is #(-2,5)#, maor axis is #2xx2sqrt5=4sqrt5# and minor axis is #2xx4=8#

Vertices are #(-2,5+-2sqrt5)# i.e. #(-2,5+-2sqrt5)# and #(-2,5+2sqrt5)#and covertices are #(-2+-4,5)# i.e. #(2,5)# and #(-6,5)#

Here we have a vertical major axis and as such eccentricity #e# and focii are #(h,k+-be)#. #e#is given by the relation #a^2=b^2(1-e^2)# i.e. #e=sqrt(1-a^2/b^2)# Here #a^2=16# and #b^2=20# and hence #e=sqrt(1-16/20)=1/sqrt5#

and focii are #(-2,5+-(2sqrt5xx1/sqrt5))# or #(-2,5+-2)# i.e. #(-2,7)# and #(-2,3)#.

graph{(5(x+2)^2+4(y-5)^2-80)((x+2)^2+(y-7)^2-0.02)((x+2)^2+(y-3)^2-0.02)=0 [-12.33, 7.67, 0.1, 10.1]}