8x^2+y^2+80x-6y+193=0
Find the center (which i think is (-5,3)
Find the major axis
Find the minor axis
&& find the distance from C to foci
come on, guy. This is just the reverse of your other post, where you had the axes and wanted the distance c.
8x^2+y^2+80x-6y+193=0
8(x^2 + 10x) + (y^2 - 6y) = - 193
8(x^2 + 10x + 25) + (y^2 - 6y + 9) = -193 + 8(25) + 9
8(x+5)^2 + (y-3)^2 = 16
(x+5)^2/2 + (y-3)^2/16 = 1
you correctly found the center: (-5,3)
semi-major axis = 4
semi-minor axis is √2
2+c^2 = 16
c = √14
extra credit: is the major axis horizontal or vertical?