Take a point P(x,y) on the parabola. The property we use is that any point on the parabola is as far from the focus F(-5,5) as it is from the directrix line. The distance from the directrix in this case is y-(-1)=y+1. The distance PF=√((x-(-5))2+(y-5)2). So PF2=(x+5)2+(y-5)2=(y+1)2.
Therefore, x2+10x+25+y2-10y+25=y2+2y+1,
x2+10x+49=12y, which must be the equation of the parabola. This can be written y=(x2+10x+49)/12.
CHECK
The vertex has the same x coordinate as the focus, that is, they both lie on the axis of symmetry, which in this case is x=-5.
Let's rewrite the equation of the parabola:
y=(x2+10x+25-25+49)/12=((x+5)2+24)/12=(x+5)2/12+2, or y-2=(x+5)2/12, which has vertex V(-5,2). V clearly lies on the axis of symmetry and shares its x-coordinate with F. Also V is 2-(-1)=3 units from the directrix and 5-2=3 units from F, putting the vertex midway between F and the directrix line. This demonstrates that y=(x+5)2/12+2 is the equation of the parabola.
