This is for review and I can't remember this.

What is the equation of a parabola with focus (6, -10) and directrix x = -2?

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- May 21st 2008, 08:19 AMAlanParabola help
This is for review and I can't remember this.

What is the equation of a parabola with focus (6, -10) and directrix x = -2? - May 21st 2008, 10:39 AMtopsquark
Personally I prefer the long way (no surprise to anyone who knows me). A parabola is defined as the locus of points such that the distance between the focus and parabola is the same as the distance between the parabola and the directrix. So

$\displaystyle \sqrt{(x - 6)^2 + (y + 10)^2} = x + 2$

-Dan - May 21st 2008, 11:10 AMAlan
The answers I have to choose from are:

(y+10)^2 = 16(x-2)

(y+4)^2 = 16(x-6)(y-10)^2 = 16(x+2)

(y-4)^2 = 16(x+6)

I'm still kind of confused... - May 21st 2008, 11:17 AMtopsquark
See this.

I prefer the definition I used because I can get equations for parabolas with axes of symmetry that aren't "nice."

In any event the directrix here is parallel to the y axis, so we can use the form

$\displaystyle (y - k)^2 = \pm 4p(x - h)$

p is the twice the distance between the directrix and the focus, so p = 4. The vertex is half way between the directrix and the focus so it is at V(2, -10). So the equation will be

$\displaystyle (y + 10)^2 = \pm 16(x - 2)$

Finally, we know that the focus is to the right of the directrix, so this parabola opens to the right. Thus we choose the + sign:

$\displaystyle (y + 10)^2 = 16(x - 2)$

Edit:

You can solve for the answer using my method:

$\displaystyle \sqrt{(x - 6)^2 + (y + 10)^2} = x + 2$

$\displaystyle (x - 6)^2 + (y + 10)^2 = (x + 2)^2$

$\displaystyle (y + 10)^2 = (x + 2)^2 - (x - 6)^2 = (x^2 + 4x + 4) - (x^2 - 12x + 36)$

$\displaystyle (y + 10)^2 = 16x - 32$

$\displaystyle (y + 10)^2 = 16(x - 2)$

-Dan - May 21st 2008, 11:49 AMAlan
I read the wiki page and tried to understand you but how do you get

$\displaystyle

(y + 10)^2 = (x + 2)^2 - (x - 6)^2 = (x^2 + 4x + 4) - (x^2 - 12x + 36)

$ - May 21st 2008, 12:12 PMmasters
http://www.mathhelpforum.com/math-he...c812335e-1.gif

Now square both sides

http://www.mathhelpforum.com/math-he...c2287097-1.gif

Transpose $\displaystyle (x-6)^2$ to the right side of the equation and expand the binomials and simplify.

http://www.mathhelpforum.com/math-he...cd214a33-1.gif

The right side simplifies to

http://www.mathhelpforum.com/math-he...a0b87b22-1.gif

Factor out 16 on the right side

http://www.mathhelpforum.com/math-he...7b96479a-1.gif