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
Sep 16, 2017

Answer:

Please see below.

Explanation:

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

whose center is #(h,k)#, #2a# and #2b# are major axis and minor axis (if #a>b#, but if #a< b#, it is reverse). In the former case major axis is parallel to #x#-axis and in latter case major axis is parallel to minor axis.

vertices are #(h,k+-b)# and #(h+-a,k)#

Eccentricity #e# can be obtained using #e=sqrt(1-("minor axis")^2/("major axis")^2)#

and focii are at a distance of #ae# on either side of center along major axis (or #be# if major axis is parallel to #y#-axis).

Hence, in #(x+5)^2/16+(y-4)^2/1=1#

center is #(-5,4)#, major axis is #2sqrt16=8#, which is parallel to #x#-axis and minor axis is #2sqrt1=2#

Vertices are #(-5,4+-1)# and #(-5+-4,4)# i.e. #(-5,5)#, #(-5,3)#, #(-1,4)# and #(-9,4)#.

Eccentricity is #e=sqrt(1-1/4)=sqrt3/2# and #ae=sqrt3#

and focii are #(-5+-sqrt3,4)# i.e. #(-5-sqrt3,4)#and #(-5+sqrt3,4)#.

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]}