How do you find the vertex, focus, and directrix of the parabola #4x-y^2-2y-33=0#?

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Answer 1 · 2017-01-18T22:56:23

Please see the explanation.

Write the given equation in #x(y) = ay^2 + bx + c# form.

#4x - y^2 - 2y - 33 = 0#

#4x = y^2 + 2y + 33#

#x(y) = 1/4y^2 + 1/2y + 33/4#

The y coordinate of the vertex, #k = -b/(2a)#:

#k = -(1/2)/(2(1/4))#

#k = -1#

The x coordinate of the vertex, #h = x(k)#:

#h = 1/4(-1)^2 + 1/2(-1) + 33/4#

#h = 8#

The vertex is the point #(8, -1)#

The focal distance is, #f = 1/(4(a))#

#f = 1/(4(1/4))#

#f = 1#

The focus is located at the point #(h + f, k)#

#(8 + 1, -1)#

The focus is the point #(9, -1)#

The equation of the directrix is #x = h - f#

#x = 8 - 1#

#x = 7#