Answer :
For any point on the graph of a parabola, the distance to the focus and the distance to the directrix are equal.
The distance to the focus:
[tex](x,y) \\ F(0,1) \\ \\ d_1=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(0-x)^2+(1-y)^2}=\\=\sqrt{(-x)^2+(1-y)^2}=\sqrt{x^2+1-2y+y^2}[/tex]
The distance to the directrix:
The directrix is a horizontal line, so the distance is the absolute value of the difference of the y-coordinates of the points.
[tex](x,y) \\ y=-1 \hbox{ so any point: } (m,-1) \\ \\ d_2=|y-(-1)|=|y+1|[/tex]
The distances are equal:
[tex]d_1=d_2 \\ \sqrt{x^2+1-2y+y^2}=|y+1| \\ (\sqrt{(x^2+1-2y+y^2})^2=|y+1|^2 \\ x^2+1-2y+y^2=(y+1)^2 \\ x^2+1-2y+y^2=y^2+2y+1 \\ x^2+1-1=y^2-y^2+2y+2y \\ x^2=4y \\ \boxed{y=\frac{1}{4}x^2}[/tex]
The distance to the focus:
[tex](x,y) \\ F(0,1) \\ \\ d_1=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(0-x)^2+(1-y)^2}=\\=\sqrt{(-x)^2+(1-y)^2}=\sqrt{x^2+1-2y+y^2}[/tex]
The distance to the directrix:
The directrix is a horizontal line, so the distance is the absolute value of the difference of the y-coordinates of the points.
[tex](x,y) \\ y=-1 \hbox{ so any point: } (m,-1) \\ \\ d_2=|y-(-1)|=|y+1|[/tex]
The distances are equal:
[tex]d_1=d_2 \\ \sqrt{x^2+1-2y+y^2}=|y+1| \\ (\sqrt{(x^2+1-2y+y^2})^2=|y+1|^2 \\ x^2+1-2y+y^2=(y+1)^2 \\ x^2+1-2y+y^2=y^2+2y+1 \\ x^2+1-1=y^2-y^2+2y+2y \\ x^2=4y \\ \boxed{y=\frac{1}{4}x^2}[/tex]
