SOLUTION: For each equation of the parabola, (a) reduce to standard and then find the (b) direction of opening, (c) vertex, (d) focus, (e) endpoints of the latus rectum, and (d) equation of

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: For each equation of the parabola, (a) reduce to standard and then find the (b) direction of opening, (c) vertex, (d) focus, (e) endpoints of the latus rectum, and (d) equation of       Log On


   

Question 1028982: For each equation of the parabola, (a) reduce to standard and then find the (b) direction of opening, (c) vertex, (d) focus, (e) endpoints of the latus rectum, and (d) equation of directrix. Graph the parabola.
x^2 – 2x – 36y + 1 = 0
(x - 〖3)〗^2 = -12 (y – 4 )
y^2 + 24x + 48 = 0
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2-2x-36y%2B1=0
Complete the square for the x if needed; but it is not needed.
x%5E2-2x%2B1=36y
%28x-1%29%5E2=36y
While not yet in standard form, this is in a form which would result IF you had derived the equation using the directrix and the focus, neither of which are given.
This is a parabola with a vertex minimum, opening upward.

If p is how far the vertex is from the focus and the directrix, then
4p=36
p=9
and
The vertex is at (1,0).
Focus (10,0)
Directrix y=-8
-
Standard form equation is y=%281%2F36%29%28x-1%29%5E2%2B0
graph%28350%2C350%2C-6%2C6%2C-6%2C6%2C%281%2F36%29%28x-1%29%5E2%29




%28x-3%29%5E2=-12%28y-4%29----
This parabola opens downward and has a vertex maximum. Again, the equation is in a manner which could come from deriving if you had been given the focus and directrix.
Read the vertex directly from this equation, since you can.
Vertex ((3,4).
I will let you figure out the directrix and focus.