how do I solve y^2=-12x

Question

How do I solve this equation? I need the Vertex, P, Directrix, Axis of Symmetry, Direction it opens, the length of the latus rectum and a graph of the parabola.

Answer 1

This is a sideways parabola, because it is y that is squared. We can write this as x=-(1/12)y^2. Its vertex is at the origin (0,0). The arms of the broad U-shaped parabola point left in the negative region of x. The axis of symmetry is the x axis. The focus is on the axis of symmetry at (0,-3), because if we put a=-(1/12), 1/4a is the distance of the focus from the vertex and 1/4a=-3. The directrix is equidistant outside the parabola at x=3. The latus rectum is where the vertical line through the focus (x=-3) meets the curve: -3=-(1/12)y^2; y=+6 so the end points are (-3,-6) and (-3,6).