SOLUTION: Find the vertex and focus of a parabola with the given equation. y2-6y-12x+21=0 and also this equation x2+4x-8y-4=0

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the vertex and focus of a parabola with the given equation. y2-6y-12x+21=0 and also this equation x2+4x-8y-4=0      Log On


   



Question 531815: Find the vertex and focus of a parabola with the given equation.
y2-6y-12x+21=0
and also this equation
x2+4x-8y-4=0

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Find the vertex and focus of a parabola with the given equation.
y2-6y-12x+21=0
and also this equation
x2+4x-8y-4=0
**
y2-6y-12x+21=0
complete the square
(y^2-6y+9)=12x-21+9=12x-12=12(x-1)
(y-3)^2=12(x-1)
This is an equation of a parabola of standard form: (y-k)^2=4p(x-h), with (h,k) being the (x,y) coordinates of the vertex. Parabola opens rightwards.
For given equation:
vertex: (1,3)
4p=12
p=3
focus: (1+p,3)=(4,3) (p units from vertex on axis of symmetry)