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?