As coefficients of #x^2# and #y^2# are positive but different, it is an ellipse and to find center, focii and vertices, let us convert it into the form #(x-h)^2/a^2+(y-k)^2/b^2=1#, where #(h,k)# is center and major and minor axis are #2a# and #2b#, (the one which is greater is major axis and smaller one is minor axis).

Vertices are #(h,k+-b)# and #(h+-a,k)# and focii are given by #(h+-ae,k)# if #a>b# and #(h,k+-be)# if #a < b#, where #e# is eccentricity given by #sqrt(1-b^2/a^2)# or #sqrt(1-a^2/b^2)#.

#16x^2+25y^2-32x+50y+16=0#

or #16x^2-32x+25y^2+50y+16=0#

or #16(x^2-2x+1-1)+25(y^2+2y+1-1)+16=0#

or #16(x-1)^2+25(y+1)^2-16-25+16=0#

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

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

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

Hence, center is #(1,-1)# and as #a>b#, major axis is #5/2# and minor axis is #2#.

Vertices are #(1,-1+-1)# and #(1+-5/4,-1)# i.e. #(1,0)#, #(1,-2)#, #(-1/4,-1)# and #(9/4,-1)#

Eccentricity is #sqrt(1-1^2/(5/4)^2)=sqrt(1-16/25)=3/5=0.6#

and #ae=5/4xx3/5=3/4# and hence focii are #(1+-3/4,-1)# i.e. #(1/4,-1)# and #(7/4,-1)#

graph{(16x^2+25y^2-32x+50y+16)((x-0.25)^2+(y+1)^2-0.002)((x-1.75)^2+(y+1)^2-0.002)((x-1)^2+y^2-0.002)((x-1)^2+(y+2)^2-0.002)((x+0.25)^2+(y+1)^2-0.002)((x-2.25)^2+(y+1)^2-0.002)((x-1)^2+(y+1)^2-0.002)=0 [-1.48, 3.52, -2.22, 0.28]}