Question

User: Convert the equation to the standard form for a hyperbola by completing the square on x and y. y^2-25x^2 + 4y + 50x - 46 = 0


a. (y+4)^2/25-(x-2)2=1
b.(x+2)^2/25-(y-1)2=1
c. (x-1)2-(x=2)^2/25=1
d.(x=2)^2/25-(x-1)2=1

Answer 1

\[y^2+4y-25(x^2-2)=46\]
\[(y+2)^2-25(x-1)^2=46+4-25=25\]
\[\frac{(y+2)^2}{25}-(x-1)^2=1\]