How do you find the center, vertices, and foci of an ellipse #(x+5)^2/144 + (y+1)^2/225 = 1#?

1 Answer
Aug 24, 2017

Answer:

Please see below.

Explanation:

#(x-h)^2/a^2+(y-k)^2/b^2=1# represents an ellipse with center #(h,k)#. Its vertices are #(h+-a,k)# and #(h,k+-b)#.

If #b>a#, then it is an ellipse with major axis parallel to #y#-axis and eccentricity is #e=sqrt(1-a^2/b^2)# and focii are #(h,k+-be)#. If #a>b#, then it is an ellipse with major axis parallel to #x#-axis and eccentricity is #e=sqrt(1-b^2/a^2)# and focii are #(h+-ae,k)#.

Hence for #(x+5)^2/144+(y+1)^2/225=1# or #(x+5)^2/12^2+(y+1)^2/15^2=1# and hence, we have ellipse with center as #(-5,-1)# and major axis parallel to #y#-axis.

Its vertices are #(-5+-12,-1)# and #(-5,-1+-15)# i.e. #(7,-1)#, #(-17,-1)#, #(-5,14)# and #(-5,-16)#.

Eccentricity is #sqrt(1-144/225)=sqrt(81/225)=9/15=3/5# and #be=15xx3/5=9#

and focii are #(-5,-1+-9)# i.e. #(-5,8)# and #(-5,-10)#.

Ellipse appears as follows:

graph{((x+5)^2/144+(y+1)^2/225-1)((x+5)^2+(y+1)^2-0.09)((x-7)^2+(y+1)^2-0.09)((x+17)^2+(y+1)^2-0.09)((x+5)^2+(y-14)^2-0.09)((x+5)^2+(y+16)^2-0.09)((x+5)^2+(y-8)^2-0.09)((x+5)^2+(y+10)^2-0.09)=0 [-40, 40, -20, 20]}