What is the equation of the quadratic graph with a focus of (3, 1) and a directrix of y = 5

Question

anonymous · Answer

which way does it open?

helder_edwin · Answer

u have the point Q(3,1) and the line L : y-5=0

use the distance formulas for P(x,y) such that
$$\large d(P,Q)=d(P,L) $$

anonymous · Answer

f(x) = one eighth (x − 3)2 + 3

 f(x) = −one eighth (x − 3)2 + 3

 f(x) = one eighth (x − 3)2 − 3

 f(x) = −one eighth (x − 3)2 −

helder_edwin · Answer

u should know this: if P(a,b) and Q(c,d) then
$$\large d(P,Q)=\sqrt{(a-c)^2+(b-d)^2} $$
and if L is the line $y=Ax+B$ then
$$\large d(P,L)=\frac{|Aa-b+B|}{\sqrt{A^2+1}} $$

whpalmer4 · Answer

Let's not make this needlessly complicated.  

Vertex form for a quadratic is

$$y=a(x-h)^2+k$$ where the vertex is at $(h,k)$ and $a$ is a constant.

We know that the parabola is the set of all points equidistant from the focus and the directrix.  That means the vertex must be on a perpendicular line drawn between the focus and the directrix.

|dw:1391287390296:dw|

Apologies for the shaky parabola!  

If the vertex is on the line x = 3, and halfway between (3,1) and (3,5), where are its coordinates?

whpalmer4 · Answer

Plug those coordinates into $$y = a(x-h)^2+k$$Compare with your answer choices.

whpalmer4 · Answer

Note that $a$ controls the vertical scale of the parabola, and whether it opens upward ($a>0$) or downward ($a<0$)

phi · Answer

for more info, see http://www.khanacademy.org/math/algebra/conic-sections/conic_parabolas/v/parabola-focus-and-directrix-1