How do I find the focus of the parabola with the equation #y=1/4x^2-3/2x+1/4#?

1 Answer
KillerBunny
Jan 21, 2015

You are in the simplest case of this kind of exercises, since you already have the parabola written in the form #y=ax^2+bx+c#.

The general rule says that, in this case, the coordinates of the focus are
#(-\frac{b}{2a}, \frac{1-\Delta}{4a})#,
where #\Delta# is the discriminant #b^2-4ac#.

Let's compute this out for your values: you have #a=\frac{1}{4}#, #b=-\frac{3}{2}#, and #c=a=\frac{1}{4}#.

So, #-\frac{b}{2a}=-frac{\frac{3}{2}}{\frac{2}{4}]=-\frac{3}{2}\frac{4}{2}=3#
and #b^2-4ac# equals #\frac{9}{4}-4\frac{1}{4}\frac{1}{4}=2#.

Thus, #\frac{1-\Delta}{4a}# equals #\frac{1-2}{4\frac{1}{4}}#, namely #-1#.

The focus of your parabola has thus coordinates #(3,-1)#

This is the graph of your parabola, you'll see that the result we got is quite reasonable
graph{x^2/4-3x/2+1/4 [-10, 10, -5, 5]}

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