x2 - 6x + 5y = -34
5y = -x^2 + 6x - 34
= -1(x^2 - 6x + 9 - 9) -34
= -(x-3)^2 + 9 - 34
y = (-1/5)(x - 3)^2 - 25/5
y = (-1/5)(x-3)^2 - 5
lots of algebraic typing ahead, so I will let you watch 4 great videos dealing with this topic by the remarkable Mr. Khan
https://www.khanacademy.org/math/precalculus/conics-precalc/focus-and-directrix-of-a-parabola/v/focus-and-directrix-introduction
the first 3 are a good introduction to your topic, and the last one deals with our question.
you should get:
http://www.wolframalpha.com/input/?i=focus+of+x%5E2+-6x%2B5y+%3D+-34
Determine the focus of the parabola with the equation x2 - 6x + 5y = -34
2 answers
moving right along, you know that the parabola
x^2 = 4py
has focus at (0,p). You have
(x-3)^2 = -5(y+5)
So, 4p = -5 and the focus is at (0,-5/4) but shifted by (3,-5) to
(3,-25/4)
as Reiny's graph shows.
x^2 = 4py
has focus at (0,p). You have
(x-3)^2 = -5(y+5)
So, 4p = -5 and the focus is at (0,-5/4) but shifted by (3,-5) to
(3,-25/4)
as Reiny's graph shows.