First we have to determine which type of a conic this is. We have only one squared term so this is a parabola. And since the y term is the one that's squared, this is a sideways-opening parabola. Opening to the right, to be specific. We need to complete the square to get this into standard form and find the coordinates of the vertex. Standard form this type of parabola is [tex](y-k)^2=4p(x-h)[/tex] where h and k are the coordinates of the vertex and p is the distance that the focus is from the vertex on the axis of symmetry. We will begin this process by setting up our equation to have the y terms on one side by themselves: [tex]y^2-14y=-12x-1[/tex]. We will now take half the linear term, square it, and add it to both sides. Our linear term is 14. Half of 4 is 7, and 7 squared is 49: [tex]y^2-14y+49=-12x-1+49[/tex]. We are going to do 2 things simultaneously now. We are going to express the left side in terms of the perfect square binomial we created in completing the square, and we are going to "do the math" on the right side. [tex](y-7)^2=-12x+48[/tex]. The left side meets the qualifications of standard form, but the right side doesn't. We need to get the x term without a coefficient, so we will factor out the -12. [tex](y-7)^2=-12(x-4)[/tex]. A couple of things are apparent to us now. The first is the coordinates of the vertex which are (4, 7), and also the value of p. Remember the "4p" that sat in front of the (x-h)...Our value sitting out front there is -12, so that means that 4p=-12 and p = -3. But p is a distance and distance will never EVER be negative so we need to absolute value of that -3, which is 3. The focus then is 3 units to the right of the vertex on the line y = 7 which is the axis of symmetry. The focus, then, is at (4+3, 4) which is (7, 4). There you go!
