standard form-conics/hyperbola- complete the square

[gymnastqueen]

okay the equation I'm given is 4x[sup:1pmy4k2g]2[/sup:1pmy4k2g]-y[sup:1pmy4k2g]2[/sup:1pmy4k2g]-8x-4y+16=0 and i'm supposed to convert it to general form for a hyperbola... i think you're supposed to do this by completing the square... I've done this, but it doesn't equal 1... when i'm finished i end up with ((x-1)[sup:1pmy4k2g]2[/sup:1pmy4k2g]/(1/4))-(y+2)[sup:1pmy4k2g]2[/sup:1pmy4k2g]=16 what do i do?? standard form is ((x-h)[sup:1pmy4k2g]2[/sup:1pmy4k2g]/a[sup:1pmy4k2g]2[/sup:1pmy4k2g])-((y-k)[sup:1pmy4k2g]2[/sup:1pmy4k2g]/b[sup:1pmy4k2g]2[/sup:1pmy4k2g])=1.... what is the best way to convert this???

 

[soroban]

Hello, gymnastqueen!

Your algebra is off . . .

\(\displaystyle 4x^2 -y^2-8x-4y+16\:=\:0\)

\(\displaystyle \text{We have: }\;4(x^2 - 2x\qquad) - (y^2 + 4y\qquad) \:=\:-16\)

. . . . . . . . \(\displaystyle 4(x^2-2x + 1) - (y^2+4y + 4) \:=\:-16 + 4 - 4\)

. . . . . . . . . . . . . . .\(\displaystyle 4(x-1)^2 - (y+2)^2 \:=\:-16\)


\(\displaystyle \text{Divide by -}16\!:\;\;\frac{4(x-1)^2}{-16} - \frac{(y+2)^2}{-16} \:=\:\frac{-16}{-16}\)

\(\displaystyle \text{And we have: }\quad \boxed{\frac{(y+2)^2}{16} - \frac{(x-1)^2}{4} \;=\;1}\)