Answer
$x^2=(-2+2\sqrt {5})y$
Work Step by Step
Equation of a parabola with vertical axis (vertex at origin): $x^2=4py$
Focus: $(0,p)$
Equation of the line:
$y=ax+b$
$y=\frac{1}{2}x+p$
Intersection point $x=2$:
$x^2=4p(\frac{1}{2}x+p)$
$2^2=4p(\frac{1}{2}·2+p)$
$4=4p+4p^2$
$0=p^2+p-1$
$p=\frac{-1±\sqrt {1^2-4(1)(-1)}}{2(1)}=\frac{-1±\sqrt {5}}{2}$
Use: $\frac{-1+\sqrt {5}}{2}$ because the parabola opens upwards.
$x^2=4(\frac{-1+\sqrt {5}}{2})y$
$x^2=(-2+2\sqrt {5})y$
