Question

*correction
The vertices of the hyperbola
25y^(2)-9x^(2)-200y+36x+139=0 are ______

Answer 1

(25y^2-200y+___) + (-9x^2 +36x +___) = -139

25(y^2-8y+(16)) -9 (x^2 -4x +(4)) = -139-36+400

25(y-4)^2 - 9(x-2)^2 = 171


25(y-4)^2 9(x-2)^2
--------- - -------- = 1
171 171

accordogng to this setup; the verts are at (2, 4 +-sqrt(171)/25)

Answer 2

(2, 4 +-sqrt(171)/5) that is; if i did it right