Question

What are the coordinates of the focus of the parabola?

Answer 1

This is a tough one.  the general form of a parabola is $(x-h) ^{2} =4p(y-k)$, 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: $- \frac{1}{16} x^{2} -x=-2$.  In order to complete the square, the leading coefficient on the squared term has to be a +1.  Ours is a $- \frac{1}{16}$, so we have to factor that out of the x terms.  When you do that you end up with $- \frac{1}{16} ( x^{2} +16x)=-2$.  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 $- \frac{1}{16}$ is still hanging out in front, which means that when we add in 64, we are actually adding in $- \frac{1}{16} *64$ which is -4.  Now here's what we have: $- \frac{1}{16} ( x^{2} +16x+64)=-2-4$which simplifies to $- \frac{1}{16}( x^{2} +16x+64)=-6$.  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 $- \frac{1}{16} (x+8) ^{2} =-6$.  Now just for simplicity, we will take baby steps.  Move the -6 back over by addition and set it back equal to y: $- \frac{1}{16}(x+8) ^{2}+6=y$.  Now we will work on getting into standard form.  Move the 6 back over by the y (baby steps, remember) to get $- \frac{1}{16} (x+8) ^{2} =y-6$.  Multiply both sides by -16 to get our "p" on the right: $(x+8) ^{2} =-16(y-6)$.  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!