hyperbolas: general equation to standarm form

[ruffnite]

Hi,
Trying to convert equation 4x^2-y^2 + 8x + 4y + 16 = 0 to standard form (x^2/a^2 - y^2/b^2 = 1)

Here is what I have done:

4x^2-y^2 + 8x + 4y + 16 = 0
(4x^2 + 8x) - (y^2 + 4y) = -16
4(x^2 + 2x) - (y^2 + 4y) = -16
4(x^2 + 2x + 1) - (y^2 + 4y + 2) = -16 + 4 - 2
4(x + 1)^2 - (y + 2)^2 = -14

I do not know how to continue from this point. Any help is appreciated.

 

[Deleted member 4993]

Hi,
Trying to convert equation 4x^2-y^2 + 8x + 4y + 16 = 0 to standard form (x^2/a^2 - y^2/b^2 = 1)

Here is what I have done:

4x^2-y^2 + 8x + 4y + 16 = 0
(4x^2 + 8x) - (y^2 - 4y) = -16
4(x^2 + 2x) - (y^2 - 4y) = -16
4(x^2 + 2x + 1) - (y^2 - 4y + 4) = -16 + 4 - 4
4(x + 1)^2 - (y - 2)^2 = -16

\(\displaystyle \frac{4(x+1)^2}{-16} \, - \, \frac{(y-2)^2}{-16} \, = \, 1\)

\(\displaystyle \frac{(y-2)^2}{16} \, - \, \frac{4(x+1)^2}{16} \, = \, 1\)

\(\displaystyle \frac{(y-2)^2}{4^2} \, - \, \frac{(x+1)^2}{2^2} \, = \, 1\) <<< Corrected - I had missed the algebraic mistakes- thanks Soroban

I do not know how to continue from this point. Any help is appreciated.

 

[soroban]

\(\displaystyle \text{Hello, ruffnite!}\)

\(\displaystyle \text{Their "standard form" is not correct.}\)
\(\displaystyle \text{They insist that the center is at the Origin.}\)

\(\displaystyle \text{Also, there are }two \text{ forms: }\;\begin{array}{ccc}\dfrac{(x-h)^2}{a^2} - \dfrac{(y-k)^2}{b^2} \;=\;1 \\ \\[-3mm] \dfrac{(y-k)^2}{b^2} - \dfrac{(x-h)^2}{a^2} \;=\;1 \end{array}\)

And you made two small errors . . .

Convert \(\displaystyle 4x^2-y^2 + 8x + 4y + 16 \:=\: 0\) to standard form.

\(\displaystyle \text{We have: }\;4x^2 + 8x - y^2 + 4y \:=\:-16\)

\(\displaystyle \text{Factor: }\;4(x^2 + 2x) - (y^2 - 4y) \:=\:-16\)
. . . . . . . . . . . . . . . . . \(\displaystyle \uparrow\)

\(\displaystyle \text{Complete the square: }\;4(x^2+2x+1) - (y^2 - 4y + 4) \:=\:-16 + 4 - 4\)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\(\displaystyle \uparrow\)

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


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

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

 

[ruffnite]

Oops. I felt something was wrong somewhere but I couldn't point it out. Thanks for the help guys =)