Tony is trying to find the equation of a quadratic that has a focus of (−1, 4) and a directrix of y = 8. Describe to Tony your preferred method for deriving the equation. Make sure you use Tony's situation as a model to help him understand.

Question

aum · Answer

A parabola has the property that any point on the parabola is equidistant from the focus and the directrix.  Sp to find the equation of the parabola (which is a quadratic) take a general point (x,y) on the parabola, find its distance from the the focus, find its distance from the directrix, equate them and simplify.

ams98 · Answer

But how do I do that with this problem specifically.

aum · Answer

Equating the distance is same as equating the square of the distance.

The square of the distance between a general point on the parabola, (x,y) and the focus, (-1, 4) is: $(x - -1)^2 + (y-4)^2 = (x+1)^2 + (y-4)^2$

The square of the distance between (x,y) and the directrix y = 8 is: $(y - 8)^2$

Equate the square of the distances:  
$(x+1)^2 + (y-4)^2 = (y-8)^2$

Simplify.

aum · Answer

$(x+1)^2 + (y-4)^2 = (y-8)^2$
$(x+1)^2 = (y-8)^2 - (y-4)^2 $
$(x+1)^2 = (y-8+y-4)(y-8-y+4) $
$(x+1)^2 = (2y-12)(-4) = 2(y-6)(-4) = -8(y-6)$
$\large \frac{-1}{8}(x+1)^2 = (y-6) $
$\large \frac{-1}{8}(x+1)^2 + 6 = y $
$\large y = \frac{-1}{8}(x+1)^2 + 6$

ams98 · Answer

holy crap is the anwser ? I was doing it SO wrong

aum · Answer

I have to double-check...

ams98 · Answer

Thank you so much! Wow

aum · Answer

Yes.  $\large y = -\frac{1}{8}(x+1)^2 + 6$  is the correct answer.