Take the 2-minute tour ×  We have a point $A(6,0)$ and a line
$k:y=2$. Show that the equation of the parabola with a locus $A$
and a directrix $k$ has the formula: $\dfrac{1}{4}x^2-3x+8$.  I
had a test on analytic geometry today and this was one of the
questions. I was 100% sure that the question was wrong, however,
all my classmates told me that they had solved it correctly
(apart from the fact that it should have been $-\dfrac{1}{4}$
instead of $\dfrac{1}{4}$, something which I noticed too,
disregard this when answering my question). So my question is
what I did incorrectly:  We use the formula of a parabola,
$y-b=\dfrac{1}{4c}(x-a)^2$. The distance between the directrix
and the locus is 2, so $c=1$. We get $y-b=\dfrac{1}{4}(x-a)^2$.
We see that the top of the parabola is $T(6,1)$, so we get
$y-1=\dfrac{1}{4c}(x-6)^2$. This can also be written as:
$y=\dfrac{1}{4}x^2 -12x + 37$. And here we go, a totally
different answer. Can anyone point out my mistakes?
share|improve this question I suppose you meant "foc