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

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

Add #y^2 + 2y + 33 to both sides:

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

Divide both side by 4:

#x = 1/4y^2 + 1/2y + 33/4" [1]"#

This type of parabola opens to left or right. Because the coefficient, a, of the #y^2# is greater than zero, we know that is opens to the right.

The vertex form of an equation of this type of parabola is:

#x = a(y - k)^2 + h#

where "a" is the coefficient of the #y^2# term and #(h, k)# is the vertex.

The focus of this type is located at #(h + 1/(4a),k)#

The equation of the directrix is #x = h - 1/(4a)#

Lets put equation [1] in vertex from. Add zero to equation [1] in the form of #1/4k^2 - 1/4k^2#:

#x = 1/4y^2 + 1/2y + 1/4k^2 - 1/4k^2 + 33/4#

Factor #1/4# from the first 3 terms:

#x = 1/4(y^2 + 2y + k^2) - 1/4k^2 + 33/4" [2]"#

Please observe that the right side of the pattern #(y - k)^2 = y^2 -2k + k^2# matches what is inside the parenthesis in equation [2]. Set the middle term of the right side of the pattern equal to the corresponding term in equation [2]:

#-2k = 2y#

#k = -1#

Substitute the left side of the pattern into the ()s in equation [2]:

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

Substitute -1 for every k:

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

Simplify the constant term:

#x = 1/4(y - -1)^2 + 8#

The vertex is at #(8, -1)#

The focus is at

#(8 + 1/(4(1/4)), -1)

This simplifies to:

(9, - 1)#

The equation of the directrix is

#x = 8 - 1/(4(1/4))#

#x = 7#