<s>[INST] Geometry: Hyperbolas I was wondering how would I complete the square for this particular hyperbola?
$4x^2 - 5y^2 + 24y = 16$
I tried this technique but to no avail:
$$4x^2 - 5(y^2 + \frac{24}{5}y) = 16$$
$$ \Rightarrow 4x^2 - 5(y + \frac{12}{5})^2 = 16 + \left(\frac{12}{5}\right)^2$$
$$\Rightarrow 4x^2 - 5(y + \frac{12}{5})^2 = \frac{544}{25} .$$
Am I doing something wrong here? On my calculator it says that the equation should be a hyperbola
 [/INST] First off, don't forget that when you factor out that "-5" to make the "completion of squares" easier, it also has to be factored out of +24y, so the equation becomes
$$4x^2 - 5(y^2 - \frac{24}{5}y ) = 16 $$
(as lab bhattacharjee already has).
You then complete the square by adding the term $\frac{12^2}{5^2}$ within the parentheses.  However, since it is inside the parentheses, what you have just "added" to the left-hand side of the equation is really  $-5 \cdot (\frac{12^2}{5^2})$, so the equation must be sustained by writing
$$4x^2 - 5(y^2 - \frac{24}{5}y  +  [\frac{12^2}{5^2}]) = 16 -5 \cdot (\frac{12^2}{5^2}) ,$$
now making the equation
$$4x^2 - 5(y -\frac{12}{5})^2 = 16 -(\frac{12^2}{5})  =  \frac{16 \cdot 5 - 144}{5} = \frac{-64}{5}  .$$
Putting the equation for this hyperbola in standard form gives us
$$\frac{-5 \cdot 4x^2}{64} - \frac{-5 \cdot 5(y -\frac{12}{5})^2}{64}   = 1 \Rightarrow \frac{25(y -\frac{12}{5})^2}{64} - \frac{20x^2}{64}    = 1.$$
[or, of course, $\frac{(y -\frac{12}{5})^2}{64/25} - \frac{x^2}{64/20}    = 1$]
So this is a "vertical" hyperbola, with its focal axis along the y-axis, since it is the y-term that is positive.  (Graphing the original and this standard-form equation confirms that they are identical.)
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