What is the focus of the parabola given by #-1/4(y+2)^2=(x-1)#?

1 Answer
Douglas K.
Nov 9, 2016

The focus is, #(0,-2)#

Explanation:

Rewrite the equation as:

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

Please observe that this is "x" as a function or "y" which means that the parabola opens either left or right (instead of up or down).
The leading coefficient, #-1/4#, determines that which direction the parabola opens; it is negative, therefore, it opens to the left.

The equation is in vertex form so its vertex is obtained by observation, #(1, -2)#

To find the signed focal distance, f, use the following equation:

#f = 1/(4a)# where "a" is the leading coefficient.

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

#f = -1#

To obtain the focus, add f to the x coordinate of the vertex, #(0,-2)#

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