This is a tough one. the general form of a parabola is
![(x-h) ^{2} =4p(y-k)](https://tex.z-dn.net/?f=%28x-h%29%20%5E%7B2%7D%20%3D4p%28y-k%29)
, where h and k are the coordinates of the vertex and p is the distance from the vertex to the focus. In order to get our parabola into this form and solve for p (which will give us our focal point), we have to complete the square. Set the parabola equal to 0, then move over the constant to get this equation:
![- \frac{1}{16} x^{2} -x=-2](https://tex.z-dn.net/?f=-%20%5Cfrac%7B1%7D%7B16%7D%20%20x%5E%7B2%7D%20-x%3D-2)
. In order to complete the square, the leading coefficient on the squared term has to be a +1. Ours is a
![- \frac{1}{16}](https://tex.z-dn.net/?f=-%20%5Cfrac%7B1%7D%7B16%7D%20)
, so we have to factor that out of the x terms. When you do that you end up with
![- \frac{1}{16} ( x^{2} +16x)=-2](https://tex.z-dn.net/?f=-%20%5Cfrac%7B1%7D%7B16%7D%20%28%20x%5E%7B2%7D%20%2B16x%29%3D-2)
. Now we can complete the square by taking half the linear term, squaring it, and adding it to both sides. Our linear term is 16, so half of 16 is 8 annd 8 squared is 64. HOWEVER, on the left side, that
![- \frac{1}{16}](https://tex.z-dn.net/?f=-%20%5Cfrac%7B1%7D%7B16%7D%20)
is still hanging out in front, which means that when we add in 64, we are actually adding in
![- \frac{1}{16} *64](https://tex.z-dn.net/?f=-%20%5Cfrac%7B1%7D%7B16%7D%20%2A64)
which is -4. Now here's what we have:
![- \frac{1}{16} ( x^{2} +16x+64)=-2-4](https://tex.z-dn.net/?f=-%20%5Cfrac%7B1%7D%7B16%7D%20%28%20x%5E%7B2%7D%20%2B16x%2B64%29%3D-2-4)
which simplifies to
![- \frac{1}{16}( x^{2} +16x+64)=-6](https://tex.z-dn.net/?f=-%20%5Cfrac%7B1%7D%7B16%7D%28%20x%5E%7B2%7D%20%2B16x%2B64%29%3D-6%20)
. Creating the perfect square binomial on the left was the point of this (to give us our vertex), so when we do that we have
![- \frac{1}{16} (x+8) ^{2} =-6](https://tex.z-dn.net/?f=-%20%5Cfrac%7B1%7D%7B16%7D%20%28x%2B8%29%20%5E%7B2%7D%20%3D-6)
. Now just for simplicity, we will take baby steps. Move the -6 back over by addition and set it back equal to y:
![- \frac{1}{16}(x+8) ^{2}+6=y](https://tex.z-dn.net/?f=-%20%5Cfrac%7B1%7D%7B16%7D%28x%2B8%29%20%5E%7B2%7D%2B6%3Dy%20%20)
. Now we will work on getting into standard form. Move the 6 back over by the y (baby steps, remember) to get
![- \frac{1}{16} (x+8) ^{2} =y-6](https://tex.z-dn.net/?f=-%20%5Cfrac%7B1%7D%7B16%7D%20%28x%2B8%29%20%5E%7B2%7D%20%3Dy-6)
. Multiply both sides by -16 to get our "p" on the right:
![(x+8) ^{2} =-16(y-6)](https://tex.z-dn.net/?f=%28x%2B8%29%20%5E%7B2%7D%20%3D-16%28y-6%29)
. We need to use our "4p" part of the standard form to find the p, which is the distance from the vertex to the focus. 4p=-16, and p = -4. That means that the focus is 4 units below the vertex. Let's figure out what the vertex is. From our equation, the vertex is ( -8, 6), and since this is an upside-down opening parabola, the focus will be aligned with the x-coordinate of the vertex. So our focus lies 4 units below 6 (6 is the y coordinate of the vertex which indicates up and down movement), so our focus has coordinates of (-8, 2), the first choice above. Told you it was a tough one! These conics are quite challenging!