how do you find the focus and the directrix line if all you have is the equation in parabolic form?
the x is not squared. i suppose you meant to type $\displaystyle y = (x - 2)^2 + 3$
now, you should know that if a parabola is in the form $\displaystyle y = a(x - h)^2 + k$, the vertex is given by $\displaystyle (h,k)$...oh, you do know that. ok, moving on.
now we want to get this parabola to look like the form i used in the post i directed you to.
the form must look like $\displaystyle (x - h)^2 = 4p(y - k)$
in this form, the focus is: $\displaystyle (h, k + p)$
and the directrix is: $\displaystyle y = k - p$
now, $\displaystyle y = (x - 2)^2 + 3$
$\displaystyle \Rightarrow (x - 2)^2 = y - 3$
the coefficient of y is 1, so where would the 4 come from? well, we can notice that $\displaystyle 1 = \frac 44 = 4 \frac 14$
thus, we have $\displaystyle (x - 2)^2 = 4 \frac 14y - 3 = 4 \frac 14(y - 3)$
so, we have $\displaystyle (x - 2)^2 = 4 \cdot \frac 14(y - 3)$ ....look familiar?
it seems you are not reading through my posts. first of all, i never said anything was k + p. neither the focus nor the directrix is k + p. the focus is a coordinate and the directrix is a horizontal line.
i am sorry, but i really do not see the problem you are having. i went through the algebraic manipulations and tried to make it plain. but let's try again.
$\displaystyle (x - {\color{red}h})^2 = 4{\color{blue}p}(y - {\color{green}k})$ ...........this is the form we want
.......$\displaystyle \downarrow$.........$\displaystyle \downarrow$.......$\displaystyle \downarrow$
$\displaystyle (x - {\color{red}2})^2 = 4 {\color{blue}\frac 14}(y - {\color{green}3})$ ............this is the form we got our equation in
now, the focus is: $\displaystyle (h, k + p)$
and the directrix is: $\displaystyle y = k - p$
can you tell me what h, p and k are?