How do you find the vertex, focus and directrix of #x^2 - 2x + 44y + 353 = 0 #?

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Answer 1 · 2017-05-16T21:39:15

Vertex: #(1, -8)#
Focus: #(1, -19)#
Directrix: #y = 3#

Start by moving the constant and #y#-value to the right side of the equation, then complete the square for #x#:

#(x^2 - 2x + 1) = -44y - 353 + 1#

#(x-1)^2 = -44y - 352#

#-1/44(x-1)^2 = y + 8#

We can now see that the vertex is #(1, -8)#

The focus is #(h, k + 1/(4a))#, which makes the focus of this equation

#(1, -8 + 1/(4(-1/44))) = (1, -19)#

The directrix of this equation is

#y = k - 1/(4a) = -8 - 1/(4(-1/44)) = 3#