What is the parabolas vertex and focus when y2 8y8x?

Answer 1

I assume you meant y2 = 8y + 8x? Subtract 8y: y2 - 8y = 8x. Complete the square: y2 - 8y + 16 = 8x + 16. Extract roots: (y - 4)2 = 8x + 16. Divide by 8: (1/8)(y - 4)2 = x + 2. So the vertex is (-2, 4). To find the focus, first consider the parabola y2/8 = x, which is nothing more than the same parabola, shifted down 4 and leftward 2. To find it's constant distance between focus and directrix, use the equation x = [1/(4p)]y2, and notice that 1/(4p) = 1/8. Solve this equation. Take the reciprocal: 4p = 8. Divide by 4: p = 2. So the distance is 2. It opens to the right, so the focus is 2 to the right of the vertex. The original parabola is the same thing, only translated, so the same thing applies - thus, the focus is 2 to the right of (-2, 4), or (0, 4).