Find the equation of the parabola 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?
 A: If the directrix is the line $y=2$:
Your value of $c$ should be negative, since the focus lies below the directrix. So, your initial equation should be $$\tag{1} y-1={-1\over 4}(x-6)^2.$$
 Apart from $c=-1$, and not $c=1$, what you had there is correct.
But, it seems you  made some algebraic errors when writing your initial equation in standard form.
To write equation $(1)$ in standard form, you could first multiply both sides by $-4$  to obtain
$$
-4y+4=(x-6)^2.
$$
Now expand the right hand side:
$$
-4y+4=x^2-12x+36.
$$
So, subtracting $4$ from both sides, we obtain $$-4y=x^2-12 x +32.$$
Finally, divide through by $-4$
$$
y={-1\over4} x^2+3x-8.
$$
But then, the proposed solution is incorrect; the right hand side should be multiplied by $-1$. (But the given answer would be correct if the directrix were the line $y=-2$.)
A: I think the sign of c depend of the position of the focus compared to the directrix (question of orientation) because if you set it at -1 you will get the answer I compute (-1/4 for 1/4).
By the way, you forgot to divide by four the affine term.
Please excuse the deplorable english of a french (thought I hope it remains understandable).
