What is the equation of the directrix and the focus of the parabola #y^2+4x-4y-8=0#?

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Answer 1 · 2017-01-20T23:20:01

When given, #x(y)=ay^2+by+c" [1]"#
#f=1/(4a)#
#k=-b/(2a)#
#h=x(k)#
The focus is #(h+f,k)#
The equation of the directrix is #x=h-f#

Write #y^2 + 4x - 4y - 8 = 0# in the form of equation [1]:

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

Matching values from equation [2] with variables in equation [1]:

#a = -1/4, b = 1, and c = 2#

Compute the value of f:

#f = 1/(4a)#

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

#f = -1#

Compute the value of k:

#k = -b/(2a)#

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

#k = 2#

Compute the value of h:

#h = x(2)#

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

#h = 3#

Compute the focus:

#(h + f, k)#

#(3 + -1, 2)#

#(2, 2)#

Compute the equation of the directrix:

#x = h - f#

#x = 3 - -1#

#x = 4#