Question

how do you write the equation of a hyperbola .... 4x^2 -y^2+16x+6y-3=0 ....... in standard form?

Answer 1

\[4x^2 -y^2+16x+6y-3=0\]\[4x^2 +16x-y^2+6y=3\]\[4x^2 +16x+16-y^2+6y-9=10\]\[4(x+2)^2-(y-3)^2=10\]\[\frac{(x+2)^2}{5/2}-\frac{(y-3)^2}{10}=1\]

Answer 2

you made mistake 6y, not 16y

Answer 3

You have \[4x^2 - y^2 + 16x + 6y - 3 = 0\] Re-arrange the terms so that you have \(x^2 + 16x\) on the left side as follows:
\[4x^2 + 16x = y^2 - 6y + 3\]\[x^2 + 4x = \frac{y^2 - 6y + 3}{4}\] Because the left side is in the form \(x^2 + bx\) add \((b/2)^2\) to both sides. In this case \((b/2)^2 = 4\) since \((4/2)^2 = (2)^2 = 4\):\[x^2 + 4x + 4 = \frac{y^2 - 6y + 3}{4} + 4\]On the left side, re-write the completed square as a perfect square binomial. On, the right side, Re-write 4 as \(\frac{16}{4}\), then combine fractions:\[(x + 2)^2 = \frac{y^2 - 6y + 3}{4} + \frac{16}{4}\]\[(x + 2)^2 = \frac{y^2 - 6y + 3 + 16}{4} \]\[(x + 2)^2 = \frac{y^2 - 6y + 19}{4} \]Multiply both sides by 4, then subtract 19 from both sides leaving \(y^2 - 6y\) on the right side:\[4(x + 2)^2 -19 = y^2 - 6y\]Again, we have the form \(y^2 + by\) so add \((b/2)^2\) to both sides. In this case, b = -6 so \((-6/2)^2 = (-3)^2 = 9\):
\[4(x + 2)^2 -19 + 9= y^2 - 6y + 9\]

Re-write the right side as a binomial square:
\[4(x+2)^2 - 10 = (y - 3)^2\]

Continue re-arranging until you reach the desired form:
\[4(x+2)^2 - (y-3)^2 = 10\]
\[\frac{4(x+2)^2 - (y-3)^2}{10} = 1\]
\[\frac{4(x+2)^2}{10} -\frac{ (y-3)^2}{10} = 1\]
\[\frac{2(x+2)^2}{5} -\frac{ (y-3)^2}{10} = 1\]