SOLUTION: Identify the vertex and directrix of the parabola (x+4)^2=-1/8(y+3) and Write the equation of a parabola a directrix at x = 1 and a focus at (-3, 0). thank you!

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Identify the vertex and directrix of the parabola (x+4)^2=-1/8(y+3) and Write the equation of a parabola a directrix at x = 1 and a focus at (-3, 0). thank you!      Log On


   

Question 481380: Identify the vertex and directrix of the parabola (x+4)^2=-1/8(y+3)
and
Write the equation of a parabola a directrix at x = 1 and a focus at (-3, 0).
thank you!
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Identify the vertex and directrix of the parabola (x+4)^2=-1/8(y+3)
and Write the equation of a parabola a directrix at x = 1 and a focus at (-3, 0).
**
(x+4)^2=-1/8(y+3)
This is an equation of a parabola of the standard form: (x-h)^2=-4p(y-k), with (h,k) being the coordinates of the vertex, and the parabola opens downwards.
For given equation:
vertex: (-4,-3)
4p=1/8
p=1/32
directrix:y=-3+1/32)
..
directrix at x = 1 and a focus at (-3, 0)
Given data shows this is an equation of a parabola of the standard form: (y-k)^2=-4p(x-h)
vertex: (-1,0)
Equation: y^2=-8(x+1)
see graph as a visual check on the answer
..
y=±(-8(x+1))^.5