Question 1044782
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Since *[tex \Large y] is the squared variable, your parabola is "sideways" and the general equation is:


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


The vertex is at the point *[tex \Large (h,k)] and the directed distances from the vertex to the focus and the directrix are *[tex \Large p\ =\ \frac{1}{4a}] and *[tex \Large p\ =\ -\frac{1}{4a}].


The axis of symmetry is the horizontal line through the vertex, so *[tex \Large y\ =\ k]


The focus is *[tex \Large (h\ +\ p,k)]  and the directrix is the vertical line *[tex \Large x\ =\ h\ -\ p]


The latus rectum is the segment with endpoints *[tex \Large (h\,+\,p,k\,+\,2p)] and *[tex \Large (h\,+\,p,k\,-\,2p)].


Now the trick is to get *[tex \Large y^2\ +\ 6x\ +\ 6y\ =\ 39] into the form *[tex \Large x\ =\ a(y\ -\ k)^2\ +\ h]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y^2\ +\ 6x\ +\ 6y\ =\ 39]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y^2\ +\ 6y\ =\ -6x\ +\ 39]


Now complete the square in the LHS:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y^2\ +\ 6y\ +\ 9\ =\ -6x\ +\ 48]


Factor both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (y\ +\ 3)^2\ =\ -6(x\ -\ 8)]


Solve for *[tex \Large x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ -\frac{1}{6}(y\ +\ 3)^2\ +\ 8]


So, by inspection you have:


*[tex \Large h\ =\ 8], *[tex \Large k\ =\ -3], and *[tex \Large a\ =\ -\frac{1}{6}\ =>\ p\ =\ -\frac{3}{2}].


You can do the rest of the arithmetic.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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