precalc
posted by 95 .
What is the transverse axis of the following equation?
25x^2+250x36y^2576y=2579

25(x^2 + 10x)  36(y^2 + 16y) = 2579
Complete the squares:
25(x^2 + 10x + 25)  25*25  36(y^2 + 16y + 64) + 36*64 = 2579
25(x+5)^2  36(y+8)^2 = 900
How convenient . . .
(x+5)^2/36  (y+8)^2/25 = 1
Looks like an hyperbola with center at (5,8) and horizontal transverse axis
y = 8. 
Thank you steve
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