SOLUTION: Find the vertex,focus,directrix and axis of symmetry in the given equation: (x+6)^2=8(-3y+17)

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Question 1041196: Find the vertex,focus,directrix and axis of symmetry in the given equation:
(x+6)^2=8(-3y+17)

Answer by AnlytcPhil(1806)   (Show Source): You can put this solution on YOUR website!



We have to get it in the standard form:



where (h,k) is the vertex, and |p| is the distance
from the vertex to the focus and also the distance
from the vertex to the directrix.  If p is positive
the parabola opens upward and if p is negative the
parabola opens downward.

The left side of



is already in that form.  We multiply out the right
side



Now we factor out -24



Reduce the fraction:



Now we can compare it to



and see that 
-h=+6 so h=-6
-k = -17/3 so k = 17/3
4p = -24, so p = -6

The vertex is (h,k) or (-6,17/3)
The parabola opens downward, since
p is a negative number.

We sketch the parabola by getting a few points
It goes through the vertex (-6,17/3) as well
as the points (-14,3), (-10,5), (-2,5), (2,3),
(10,-5)


 
The focus (which is inside the parabola is |p| or 17/6
units below the vertex.  It has the same x-coordinate -6
So to find its y-coordinate we subtract 6 from the
y-coordinate of the vertex 17/3:

  

So the focus is (-6,-1/3)



Finally we will draw the directrix, which is a horizontal
line (in green below) which is |p| = 6 units outside (above)
the vertex.  We determine how far above the x-axis that is
by adding 6 to the y-coordinate of the vertex 17/3:



So the directrix is a horizontal line y = 35/3



  
The axis of symmetry is the vertical line (in blue below) 
through both the vertex and the focus which cuts the
parabola in two.  It is the line x=(the x-coordinate of
the vertex and focus).  In this case the equation of the
axis of symmetry is x=-6:


  
Edwin

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