Question

what would be the standard form for this hyperbola equation? 4y 2 − x 2 − 24y − 4x + 16 = 0

Answer 1

just complete the sqaures

Answer 2

(4y^2-24y) - (x^2 +4x) +16 =0

Answer 3

4(y^2 -6y) -(x^2+4x) +16 =0

4(y^2 -6y+9 -9) - (x^2 +4x +4 -4) +16 =0

Answer 4

4 [ (y-3)^2 -9] - ( (x+2)^2 -4) +16 =0

Answer 5

so \[4(y-3)^2 - 36 -(x+2)^2 +4 +16 =0 \]

Answer 6

so is the standard form (y-3)^2/9-(x-4)^2/16=1?

Answer 7

or x^2/16-y^2/4=1?

Answer 8

\[4(y-3)^2 -(x+2)^2 = 16\]

Answer 9

then

Answer 10

\[\frac{(y-3)^2}{4}- \frac{(x+2)^2}{16} =1\]