Question 1086398
The equation of the parabola can be written  {{{ y = (1/8)x^2 }}} 
Graphed below (green line is latus rectum):
—


{{{ graph( 400, 400, -4, 4, -5, 5, (1/8)*x^2, y=2)
 }}}

—

For a (vertical) parabola with vertex @(h,k):   {{{ 4p(y-k) = (x-h)^2 }}} 
Vertex is at (0,0) so h=k=0:   so this parabola has equation  {{{ y = (1/8)x^2 }}}

Solve for p (the distance from vertex to focus) by comparison:  {{{ y = x^2/(4p) }}} ==>  {{{ 4p = 8 }}} ==> p=2
So focus is at (0,2).   Latus rectum passed through (0,2) and intersects parabola where {{{y = 2 = (1/8)x^2 }}} or  (-4,2) and (4,2)

Using just the first quadrant & symmetry with 2nd quadrant:
Area = {{{ 2* int(( 2 - (1/8)x^2), dx, 0, 4) }}}
         = {{{ 2* (2x - (1/24)x^3) }}}  evaluated at 4 and 0
          = {{{ 2*((8-(1/24)(64)) - (0-0)) }}}
          = {{{ highlight(32/3) }}}