Using a directrix of y = -2 and a focus of (1, 6), what quadratic function is created?
f(x) = 1/8 (x - 1)^2 - 2
f(x) = -1/8 (x + 1)^2 - 2
f(x) = -1/16 (x + 1)^2 - 2
f(x) = 1/16 (x - 1)^2 + 2

I think it's the first?

Question

Using a directrix of y = -2 and a focus of (1, 6), what quadratic function is created?
 f(x) = 1/8 (x - 1)^2 - 2
 f(x) = -1/8 (x + 1)^2 - 2
 f(x) = -1/16 (x + 1)^2 - 2
 f(x) = 1/16 (x - 1)^2 + 2

I think it's the first?

anonymous · Answer

It is the last one.
Actually we can find the equation by equation the distances from a point (x,y) to the focus and to the directirx.

anonymous · Answer

Let me show you how
$$
(y+2)^2=(x-1)^2+(y-6)^2\
\text { after some simplification you get}\
-x^2+2 x-33=-16 y\
y=\frac{1}{16} \left(x^2-2
   x+33\right)=\frac{1}{16} \left(x^2-2
   x+1 +32\right)=\
\frac{1}{16} \left((x-1)^2 
  +32\right)=\frac{1}{16} (x-1)^2 
+2
$$

anonymous · Answer

Thank you very much ! (:

anonymous · Answer

YW