Question

Find an equation for the hyperbola in standard form.
25x^2-16y^2+250x-224y-559=0

a. (x+5)^2/25-(y+7)^2/16=1
b. (x+7)^2/25-(y+5)^2/16=1
c. (x+7)^2/16-(y+5)^2/25=1
d. (x+5)^2/16-(y+7)^2/25=1
e. none of the above

Answer 1

i have to start over the 16 should be a 14 on the y-variables

Answer 2

factor the x and y variables to that the squared terms have a coefficient of +1, then complete the squre on each variable
\[25(x^2+10x+25)−16(y^2+14y+49)=559+625−784\]factor as a perfect square trinomial\[25(x+5)^2-16(y+8)^2=400\]now divide through by 400\[\frac{(x+5)^2}{16}-\frac{(y+7)^2}{25}=1\]Answer D.