Determine the focus of the parabola with the equation x2 - 6x + 5y = -34

2 answers

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
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.
Similar Questions
  1. determine vertex, focus and directrix of parabola. then graph the parabola.first equation: y^2 -2x-8=0 second equation: y^2 -4y
    1. answers icon 1 answer
    1. answers icon 0 answers
  2. determine vertex, focus and directrix of parabola. then graph the parabola.first equation: y^2 -2x-8=0 second equation: y^2 -4y
    1. answers icon 1 answer
  3. determine vertex, focus and directrix of parabola. then graph the parabola.second equation: y^2 -4y -12x= 8 Please help me this
    1. answers icon 1 answer
more similar questions