# Math Help - Locus Help

1. You can move them, but you'll get a different parabola. The parabola I posted has it's specific focus and directrix. As does any parabola.
You can't just move the focus and directrix around and keep the same parabola.

Let's find the equatio of a parabola with vertex at V(-4,2) and directrix at y=5.

See, the distance from the vertex to the directrix(line y=5) is 3 units. 5-2=3.

So, the focus is F(-4,-1). Because 2-3=-1.

We can use $(x-h)^{2}=4p(y-k)$

We know h and k, the coordinates of the vertex. So, $(x+4)^{2}=4(-3)(y-2)$

$(x+4)^{2}=-12(y-2)$

We want to express it as $ax^{2}+bx+x$

$x^{2}+8x+16=-12y+24$

$12y=-x^{2}-8x+8$

$y=\frac{-1}{12}x^{2}-\frac{2}{3}x+\frac{2}{3}$

There she is.

Parabolas are used a lot in real world applications. One area is in highway construction. Vertical curves on roads(hills and valleys) are laid out using the principles of parabolas.

2. Originally Posted by galactus
You can move them, but you'll get a different parabola. The parabola I posted has it's specific focus and directrix. As does any parabola.
You can't just move the focus and directrix around and keep the same parabola.

Parabolas are used a lot in real world applications. One area is in highway construction. Vertical curves on roads(hills and valleys) are laid out using the principles of parabolas.
Ah, interesting. And how do we determine the focus of a parabola? What's the formula?

3. I just showed. It's the same distance as from the vertex to the directrix.

$\frac{1}{4a}=p$. |p| is the distance from the vertex to the focus.

'a' is the coefficient of the x^2. Remember, we had -1/6 on the one I graphed. So, $\frac{1}{4(-1/6)}=|\frac{3}{2}|$

It's 3/2 units from the focus to the vertex or from the vertex to the directrix.

If p<0, then the parabola opens down. If p>0, then it opens up.

Of course, they can be horizontal too. In that case, you have

$x=\frac{1}{4p}y^{2}$ instead of $y=\frac{1}{4p}x^{2}$

If it's a horizontal parabola, then if p>0 it opens to the right and if p<0 it opens left.

4. Originally Posted by galactus
I just showed. It's the same distance as from the vertex to the directrix.

$\frac{1}{4a}=p$. |p| is the distance from the vertex to the focus.

'a' is the coefficient of the x^2. Remember, we had -1/6 on the one I graphed. So, $\frac{1}{4(-1/6)}=|\frac{3}{2}|$

It's 3/2 units from the focus to the vertex or from the vertex to the directrix.

If p<0, then the parabola opens down. If p>0, then it opens up.

Of course, they can be horizontal too. In that case, you have

$x=\frac{1}{4p}y^{2}$ instead of $y=\frac{1}{4p}x^{2}$

If it's a horizontal parabola, then if p>0 it opens to the right and if p<0 it opens left.
Am i correct then when i say that the x-coordinate of the focus is equal to the symmetry axis, and the y-coordinate is equal to that of the directrix?

$Ax^2+Bxy+Cy^2+De+Fy+G=0$.