I need help finding vertex,foci, and directrix of this parabola. Thanks
x^2+5x-4y-1=0
y = (1/4)(x^2 + 5x - 1)
Complete the square inside the parenthesis:
y = (1/4)([x^2 + 5x + (5/2)^2] - (5/2) - 1)
y = (1/4)(x + 5/2)^2 - (1/4)(7/2)
y = (1/4)(x + 5/2)^2 - 7/8
4(y + 7/32) = (x + 5/2)^2
The general form for an upward opening parabola with a vertex at (h, k) is:
4p(y - k) = (x - h)^2
where p is the distance of the vertex to the focus, or the distance of the vertex to the directrix.
So the vertex of your parabola is
(h, k) = (-5/2, -7/32)
The focus is p units directly above the vertex, and p = 1 (according to the above form) so the focus is at
(-5/2, -7/32 + 1) = (-5/2, 25/32)
The directrix is, in this case, a horizontal line p = 1 units below the vertex, so this line is
y = -7/32 - 1 = -39/32
-Dan