Question 1041166
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Complete the square on *[tex \Large y] and rearrange into standard conics form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  4p(x\ -\ h)\ =\ (y\ -\ k)^2]


Where *[tex \Large (h,k)] is the vertex and *[tex \Large p] is the distance from the vertex to the focus and from the vertex to the directrix.  The *[tex \Large y] term is squared so this parabola has a horizontal axis of symmetry and a vertical directrix.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  y^2\ -\ 5x\ +\ 12y\ =\ -16]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  -5x\ =\ -y^2\ -\ 12y\ -\ 16]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  5x\ -\ 16\ +\ 36\ =\ y^2\ +\ 12y\ +\ 36]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  5\left(x\ +\ 4\right)\ =\ (y\ +\ 6)^2]


Hence, vertex at *[tex \Large \left(-4,\,-6\right)], *[tex \Large 4p\ =\ 5] so *[tex \Large p\ =\ \frac{5}{4}], directrix at *[tex \Large x\ =\ -4\ -\ \frac{5}{4}], focus *[tex \Large \left(-4\ +\ \frac{5}{4},-6\right)], and axis of symmetry *[tex \Large y\ =\ -6]


You can do your own arithmetic to simplify the expressions for the directrix and the focus.

*[illustration horizParabolavertfocdir_sym.jpg].


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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