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Please help me solve this problem: \"Determine the equation of the parabola with the focus at (-4, 4) and where the directrix is the line y=-2. Find the two points that define the latus rectum and sketch the parabola.
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\n" ); document.write( "Equation is that of a parabola which opens upwards of the standard form: (x-h)^2=4p(y-k), (h,k) being the (x,y) coordinates of the vertex.(note that the directrix, y=-2, is below the y-coordinate, 4, of given focus)
\n" ); document.write( "Axis of symmetry:x=-4
\n" ); document.write( "y-coordinate of vertex=(4+(-2))/2=2/2=1 (midpoint formula)
\n" ); document.write( "x-coordinate=-4
\n" ); document.write( "vertex:(-4,1)
\n" ); document.write( "p=distance from vertex to directrix or focus on the axis of symmetry=3
\n" ); document.write( "4p=12
\n" ); document.write( "Equation: (x+4)^2=12(y-1)
\n" ); document.write( "..
\n" ); document.write( "latus rectum: a line passing thru the focal point perpendicular to the axis of symmetry connecting to both sides of the parabola at the following points:
\n" ); document.write( "y-coordinate for both points=4 (same as that of focus)
\n" ); document.write( "solving for x-coordinates:
\n" ); document.write( "(x+4)^2=12(y-1)
\n" ); document.write( "(x+4)^2=12(4-1)
\n" ); document.write( "(x+4)^2=36
\n" ); document.write( "x+4=±√36=±6
\n" ); document.write( "x=±6-4
\n" ); document.write( "x=-10
\n" ); document.write( "or
\n" ); document.write( "x=1
\n" ); document.write( "end points of latus rectum: (-10,4) and (2,4)
\n" ); document.write( "..
\n" ); document.write( "see graph below:
\n" ); document.write( "y=((x+4)^2+12)/12
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