1. confusing math on conics

Given (y-2)^2 - X^2/4=1, do the following

a) write the equation in general form:

(y-2)^2 - X^2/4=4
4(y-2)^2-1(x)^2=4
4(y^2-4y+4) -1(x^2+x)=4
4y^2-16y+16-(x^2+x)= 4
4y^2-16y+16-X^2+x=4
-x^2+4y^2+x-16y-16=4
-x^2+4y^2+x-16y-20=0

general form:
-x^2+4y^2+x-16y-20=0

How do I locate the vertices and center of this hyperbola??

2. Hello, Justin!

Given: . $(y-2)^2 - \frac{x^2}{4}\:=\:1$

1) Find the center and vertices of the hyperbola.

We have: . $\frac{(y-2)^2}{1} - \frac{x^2}{4}\:=\:1$

The center is: $(0,2)$. . $a = 2,\:b = 1$

The hyperbola is "vertical": $\begin{array}{ccc}\lor \\ \land\end{array}$

The vertices are one unit above and one unit below the center.
. . $V\!: \;(0,3),\;(0,1)$

2) Write the equation in general form

Your algebra is faulty . . .

We have: . $\frac{(y-2)^2}{1} - \frac{x^2}{4}\:=\:1$

Multiply by $4\!:\;\;4(y-2)^2 - x^2\;=\;4$

. . $4(y^2 - 4y + 4) - x^2\;=\;4$

. . $4y^2 - 16y + 16 - x^2\:=\:4$

. . $-x^2 + 4y^2 - 16y + 12\;=\;0\;\;\text{ or }\;\;x^2 - 4y^2 + 16y - 12\;=\;0$