SOLUTION: convert the equation to standard for by completing the square on x. Then find the vertex, focus and directrix of the parabola. then graph x^2+2x-8y-31=0

Algebra ->  Trigonometry-basics -> SOLUTION: convert the equation to standard for by completing the square on x. Then find the vertex, focus and directrix of the parabola. then graph x^2+2x-8y-31=0       Log On


   



Question 1089163: convert the equation to standard for by completing the square on x. Then find the vertex, focus and directrix of the parabola. then graph
x^2+2x-8y-31=0

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2%2B2x-8y-31=0
The "vertex" form of a parabola with its vertex at (h,+k) is:
regular: y+=+a%28x-h%29%5E2+%2B+k
sideways: x+=+a%28y+-k%29%5E2+%2B+h+
The conics form of the parabola equation (the one you'll find in advanced or older texts) is:
regular: 4p%28y+-k%29+=+%28x-h%29%5E2
sideways: 4p%28x+-h%29+=+%28y+-k%29%5E2
where the value of 4p is actually the same as the value of 1%2F4a

%28x%5E2%2B2x%2Bb%5E2%29-b%5E2-31=8y....since coefficients a=1 and 2ab=2, we have 2%2A1%2Ab=2->2b=2->b=1
8y=%28x%5E2%2B2x%2B1%5E2%29-1%5E2-31
8y=%28x%2B1%29%5E2-1-31
y=%281%2F8%29%28x%2B1%29%5E2-32%2F8
y=%281%2F8%29%28x%2B1%29%5E2-4=> h=-1 and k=-4, and
the vertex is at (-1,-4)
The focus is "p" units from the vertex:
p+=1%2F4a=1%2F%284%281%2F8%29%29=2,
so p=2
then,
the focus is 2 unit above the vertex, at (-1, -2),
and the directrix is the horizontal line+y+=+-6, p or two+units below the vertex