Answer :
[tex](x;\ y)-the\ coordinates\ any\ point\ on\ the \ parabola.\\\\Distance\ between\ the\ point\ (x;\ y)\ and\ the\ focus\ (-5;\ 5)\\and\ distance\ between\ the\ point\ (x;\ y)\ and\ a\ direct\ are\ equal.\\\\Distance\ between\ (x;\ y)\ and\ (-5;\ 5):\\\sqrt{(x-(-5))^2+(y-5)^2}=\sqrt{(x+5)^2+(y-5)^2}\\\\Distance\ between\ (x;\ y)\ and\ y=-1:\\|y-(-1)|=|y+1|\\\\We\ equate\ the\ two\ expressions:\\\sqrt{(x+5)^2+(y-5)^2}=|y+1|[/tex]
[tex]Square\ both\ sides:\\(x+5)^2+(y-5)^2=(y+1)^2\\\\Use:(a\pm b)^2=a^2\pm2ab+b^2\\\\x^2+2\cdot x\cdot5+5^2+y^2-2\cdot y\cdot5+5^2=y^2+2\cdot y\cdot1+1^2\\x^2+10x+25+\not y^2-10y+25=\not y^2+2y+1\\x^2+10x-10y+50=2y+1\\2y+1=x^2+10x-10y+50\ \ \ \ |subtract\ 1\ from\ both\ sides\\2y=x^2+10x-10y+49\ \ \ \ |add\ 10y\ to\ both\ sides\\12y=x^2+10x+49\ \ \ \ |divide\ both\ sides\ by\ 12\\\\\boxed{y=\frac{1}{12}x^2+\frac{5}{6}x+\frac{49}{12}}[/tex]
[tex]Square\ both\ sides:\\(x+5)^2+(y-5)^2=(y+1)^2\\\\Use:(a\pm b)^2=a^2\pm2ab+b^2\\\\x^2+2\cdot x\cdot5+5^2+y^2-2\cdot y\cdot5+5^2=y^2+2\cdot y\cdot1+1^2\\x^2+10x+25+\not y^2-10y+25=\not y^2+2y+1\\x^2+10x-10y+50=2y+1\\2y+1=x^2+10x-10y+50\ \ \ \ |subtract\ 1\ from\ both\ sides\\2y=x^2+10x-10y+49\ \ \ \ |add\ 10y\ to\ both\ sides\\12y=x^2+10x+49\ \ \ \ |divide\ both\ sides\ by\ 12\\\\\boxed{y=\frac{1}{12}x^2+\frac{5}{6}x+\frac{49}{12}}[/tex]
