Answer:
[tex]\displaystyle y=\frac{1}{4}x^2[/tex]
Step-by-step explanation:
Let (x, y) be a point on the parabola.
By definition, any point on the parabola is equidistant from the focus and the directrix
The distance from the focus is given by:
[tex]\begin{aligned} d&=\sqrt{(x-0)^2+(y-1)^2\\\\&=\sqrt{x^2+(y-1)^2}\end{aligned}[/tex]
The distance from the directrix is given by:
[tex]d=|y-(-1)|=|y+1|\text{ or } |-1-y|[/tex]
So:
[tex]\sqrt{x^2+(y-1)^2}=|y+1|^[/tex]
Square both sides. Since anything squared is positive, we can remove the absolute value:
[tex]x^2+(y-1)^2=(y+1)^2[/tex]
Square:
[tex]x^2+(y^2-2y+1)=y^2+2y+1[/tex]
Hence:
[tex]x^2=4y[/tex]
So, our equation is:
[tex]\displaystyle y=\frac{1}{4}x^2[/tex]
