If the vertex is on the line x=2; axis parallel to x axis, Latus rectum = 6 and passing through (8,2)
Help please. No idea. The axis parallel to x axis made it confusing.
1. Since the point P(8, 2) lies at the right side of the vertex the parabola opens to the right. The general equation of such a parabola is
$\displaystyle (y-h)^2=4p(x-k)$ with the vertex at V(k, h)
2. You already know that the latus rectum l = 4p which is 6 in your question.
3. You know the x-coordiante of the vertex (k = 2) and you know the coordinates of one point of the parabola. Plug in all known values:
$\displaystyle (y-h)^2=6(x-2)$
Solve for h.
EDIT: There are two parabolas which satisfy the given conditions!