What is the center, the​ vertices, and the foci of the ellipse? 16x^2+25y^2-160x+200y+ 400=0

1 Answer
Apr 17, 2018

Center of ellipse is #(5,-4)#; vertices are #(0,-4)#, #(10,-4)#, #(5,0)# and #(5,-8)# and focii are #(2,-4)# and #(8,-4)#.

Explanation:

There is no #xy# term here, so tha ellipse has its major and minor axis parallel to #x# and #y# axes and hence we can write the equation in form

#(x-h)^2/a^2+(y-k)^2/b^2=1#, whose center is #(h,k)# and major and minor axes are #2a# and #2b# (if #a>b#, otherwise reverse i.e. minor axis is #2a# and major axis is#2b#). Focii are always along major axis and here at a distance of #ae# (or #be# if #2b# is major axis) on either side of center, where Let us convert the equation accordingly

#16x^2+25y^2-160x+200y+400=0# can be written as

#16(x^2-10x)+25(y^2+8y)+400=0#

or #16(x^2-10x+25)+25(y^2+8y+16)-400-400+400=0#

or #16(x-5)^2+25(y+4)^2=400#

or #(x-5)^2/25+(y+4)^2/16=1#

or #(x-5)^2/5^2+(y+4)^2/4^2=1#

Hence, center of ellipse is #(5,-4)# and major axis is #10# and minor axis is #8#.

Verices will be #5# units on either side of center parallel to #x#-axis and #4# units on either side of center parallel to #y#-axis

i.e. vertices are #(0,-4)#, #(10,-4)#, #(5,0)# and #(5,-8)#.

Eccentricity of ellipse is #sqrt(1-b^2/a^2)=sqrt(1-16/25)=3/5#

and hence focii are #5xx3/5=3# units on either side of #(5,-4)# i.e. focii are at #(2,-4)# and #(8,-4)#.

graph{((x-5)^2+y^2-0.03)((x-10)^2+(y+4)^2-0.03)((x-5)^2+(y+8)^2-0.03)(x^2+(y+4)^2-0.03)((x-5)^2+(y+4)^2-0.03)((x-8)^2+(y+4)^2-0.03)((x-2)^2+(y+4)^2-0.03)(16x^2+25y^2-160x+200y+400)=0 [-5.83, 14.17, -8.72, 1.28]}