How do I find the latus rectum of a parabola?

Question

I was asked to find the two points that define the latus rectum of the parabola of the equation x² = 9/2y. It also has a vertex at (0,0) and contains the point (3,2).

Answer 1

To answer this question, we need to first determine the location of the focus of the parabola. In general terms, if the equation of a parabola is:

y = p(x-r)² + s

then the focus is at the point (r, (s + 1/4p))

In this case, the equation of the parabola is (9/2)y=x² ; rearranging we get:

y = (2/9)x² as the equation of the parabola. Comparing with the general equation above, we see that r=0 and s=0 in this case. Hence, the focus is at (0, (0+1/(4*(2/9))) i.e. (0, 9/8).

Next, we need to find the latus rectum. By definition, this is the line perpendicular to the axis of the parabola, passing through the focus and intersecting at either end with the parabola.

Since the axis of the parabola here is the y-axis, the latus rectum must be parallel to the x-axis. Further, since the focus is at (0, 9/8), the latus rectum is along the line y=9/8 (which is parallel to the x-axis).

The last step is to find the points of intersection of the latus rectum with the parabola.

Using the equation of the parabola (9/2)y=x² and substituting y=9/8, we solve for x as x=sqrt(9²/16) which yields: x = +/- 3/2. Thus the latus rectum is the line from (-3/2, 9/8) to (3/2, 9/8).

Note: The length of the latus rectum will be the length from x=-3/2 to the y-axis plus the length from the y-axis to x=3/2. That is 2*(3/2) =2.