Question 1129284

Find the latus rectum, vertex and directrix of the parabola y^2-8x-2y+17=0
<pre>This is a parabola with a HORIZONTAL AXIS of SYMMETRY, and so, its equation is: {{{matrix(1,3, (y - k)^2, "=", 4p(x - h))}}}
Given equation: {{{matrix(1,3, y^2 - 8x - 2y + 17, "=", 0)}}}
After COMPLETING the SQUARE on y, we get: {{{matrix(1,7, (y - 1)^2, "=", 8x - 16, "======>", (y - 1)^2, "=", 8(x - 2))}}}
Comparing that to the equation of a parabola with a HORIZONTAL AXIS of SYMMETRY, or {{{matrix(1,3, (y - k)^2, "=", 4p(x - h))}}}, we see that:
h = 2; k = 1			
4p = 8, or {{{matrix(1,5, p, "=", 8/4, "=", 2)}}}
{{{highlight_green(system(matrix(1,9, Latus, "Rectum:",	"4|p|", "=", "4|2|", "=", 4(2), "=", 8), matrix(1,6, "Vertex:", "(h,", "k)", "=", "(2,", "1)"), matrix(1,10, "Directrix:", x, "=", h - p, "======>", x, "=", 2 - 2, "=", 0)))}}}