Analytic Geometry - Parabola 2

A parabola in the form of $x^2 = 4y$ has a focus of $F(0, 1)$ and a directrix of $y = -1$ .

Let the $x$ -coordinate of $A$ be $p$ . Since $A$ is on $x^2 = 4y$ , its coordinates are $A(p, \frac{p^2}{4})$ .

Line $AF$ then has a slope of $m_{AF} = \frac{p^2 - 4}{4p}$ .

Since $BF$ is perpendicular to $AF$ , it has a slope of $m_{BF} = -\frac{4p}{p^2 - 4}$ .

Since $BF$ goes through $F(0, 1)$ , its equation is $y = -\frac{4p}{p^2 - 4}x + 1$ .

Since $B$ is on $y = -1$ and $y = -\frac{4p}{p^2 - 4}x + 1$ , its coordinates are $B(\frac{p^2 - 4}{2p}, -1)$ .

Line $AB$ then has a slope of $m_{AB} = \frac{p}{2}$ .

The tangent line to $x^2 = 4y$ at $x = p$ also has a slope of $y' = \frac{p}{2}$ . Therefore, $AB$ is tangent to the parabola, so $X = 1$ .

Since $M$ and $N$ are on $x^2 = 4y$ and $y = -\frac{4p}{p^2 - 4}x + 1$ , their coordinates are $M(\frac{2(p - 2)}{p + 2}, \frac{(p - 2)^2}{(p + 2)^2})$ and $N(\frac{-2(p + 2)}{p - 2}, \frac{(p + 2)^2}{(p - 2)^2})$ .

Line $AM$ then has a slope of $m_{AM} = \frac{p^2 + 4p - 4}{4p + 8}$ and line $AN$ then has a slope of $m_{AN} = \frac{p^2 - 4p - 4}{4p - 8}$ .

For $AM \perp AN$ , $m_{AM} \cdot m_{AN} = -1$ , or $(\frac{p^2 + 4p - 4}{4p + 8})(\frac{p^2 - 4p - 4}{4p - 8}) = -1$ . This simplifies to $(p^2 - 12)(p^2 + 4) = 0$ , whose only positive real solution (for the first quadrant) is $p = 2\sqrt{3}$ .

Therefore, the coordinates of $A$ are $A(p, \frac{p^2}{4}) = A(2\sqrt{3}, 3)$ , which means $Y = y - x = 3 - 2\sqrt{3}$ .

Therefore, $\lfloor 10000(X + Y) \rfloor = \lfloor 10000(1 + 3 - 2\sqrt{3}) \rfloor = \boxed{5358}$ .