SOLUTION: Find the coordinates of the vertex,focus,ends of the latus rectum and the equation of the directrix.Draw the parabola of: x^2=12(y+7)

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the coordinates of the vertex,focus,ends of the latus rectum and the equation of the directrix.Draw the parabola of: x^2=12(y+7)      Log On


   



Question 719601: Find the coordinates of the vertex,focus,ends of the latus rectum and the equation of the directrix.Draw the parabola of:
x^2=12(y+7)

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Find the coordinates of the vertex,focus,ends of the latus rectum and the equation of the directrix.Draw the parabola of:
x^2=12(y+7)
...
This is an equation of a parabola that opens upwards:
Its basic form of equation:(x-h)^2=4p(y-k)
For given equation: x^2=12(y+7)
vertex: (0,-7)
axis of symmetry: x=0
4p=12
p=3
focus: (0,-4) (p-units above vertex on the axis of symmetry)
Ends of latus rectum(focal width):
plugs in y-coordinate of focus(-4) then solve for x
x^2=12(y+7)
x^2=12(-4+7)=12*3=36
x=±√36=±6
ends of latus rectum: (-6,-4) and (6,-4)
see graph below:
+graph%28+300%2C+300%2C+-10%2C+10%2C+-10%2C+10%2C+%28x%5E2-84%29%2F12%29