Abs' question at Yahoo! Answers regarding finding a parabola given the focus and directrix

[MarkFL]

Here is the question:

Find the formula of this parabola?


Derive the equation of the parabola with a focus at (-5, -5) and a directrix of y = 7.

So... I've tried this one over and over but can't seem to get the right answer. Help anyone?

I have posted a link there to this topic so the OP can see my work.

 

[MarkFL]

Hello Abs,

A parabola is defined as the locus of all points $(x,y)$ equidistant from a point (the focus) and a line (the directrix). Using the square of the distance formula, we may write:

\(\displaystyle (x+5)^2+(y+5)^2=(y-7)^2\)

\(\displaystyle x^2+10x+25+y^2+10y+25=y^2-14y+49\)

Combining like terms, we obtain:

\(\displaystyle x^2+10x+1+24y=0\)

Solving for $y$, we get the quadratic function:

\(\displaystyle y=-\frac{x^2+10x+1}{24}\)

 

[soroban]

Hello, Abs!

Find the equation of the parabola with focus at (-5, -5)
and directrix [tex]y = 7. [/tex]

                    |
                    |7
          - - . - - + - - -
              :     |
              :V    |
              o     |
    - - - * - : - * + - - - - -
        *     :     *
       *      o     |*
              :F    |
      *       :     | *
                    |

The focus [tex](F)[/tex] is (-5,-5).
The vertex [tex](V)[/tex] is (-5,1).

The form of this parabola is: [tex](x-h)^2 \:=\:4p(y-k)[/tex]
where [tex](h,k)[/tex] is the vertex,
and [tex]p[/tex] is the directed distance from [tex]V[/tex] to [tex]F.[/tex]

We have: [tex](h,k) = (-5,1)[/tex] and [tex]p = -6.[/tex]

The equation is: .[tex](x+5)^2 \:=\:-24(y-1)[/tex]