parabola problem

To solve for the coordinates of point $A$, let’s first analyze the properties of the given parabola and the geometry involved.

The equation of the parabola is given by

$$x^2 = 4y.$$

The focus $F$ of this parabola is at $(0, 1)$. We can derive the directrix of the parabola from its standard form, which implies:

- The distance from any point $(x, y)$ on the parabola to the focus is equal to its distance to the directrix line $y = -1$.

Next, let’s consider a line passing through the focus $F$ with a slope $m$:

$$y - 1 = m(x - 0) \quad \Rightarrow \quad y = mx + 1.$$

We now substitute this expression for $y$ into the parabola equation to find the intersection points $M$ and $N$.

Substituting $y = mx + 1$ into $x^2 = 4y$, we get:

$$x^2 = 4(mx + 1).$$

This results in the quadratic equation:

$$x^2 - 4mx - 4 = 0.$$

Using the quadratic formula, we can find the $x$-coordinates of points $M$ and $N$:

$$x = \frac{4m \pm \sqrt{(4m)^2 + 16}}{2} = 2m \pm 2\sqrt{m^2 + 1}.$$

This corresponds to $x_M = 2m + 2\sqrt{m^2 + 1}$ and $x_N = 2m - 2\sqrt{m^2 + 1}$.

Next, we can determine the corresponding $y$-coordinates of points $M$ and $N$:

For $M$:

$$y_M = m(2m + 2\sqrt{m^2 + 1}) + 1 = 2m^2 + 2m\sqrt{m^2 + 1} + 1,$$

For $N$:

$$y_N = m(2m - 2\sqrt{m^2 + 1}) + 1 = 2m^2 - 2m\sqrt{m^2 + 1} + 1.$$

Now, we want to find $A$ such that angle $MAN = 90^\circ$, meaning line $FA$ is perpendicular to line $MN$.

To do this, we firstly determine the slope $m_{MN}$ of line segment $MN$:

$$m_{MN} = \frac{y_M - y_N}{x_M - x_N} = \frac{(2m^2 + 2m\sqrt{m^2 + 1} + 1) - (2m^2 - 2m\sqrt{m^2 + 1} + 1)}{(2m + 2\sqrt{m^2 + 1}) - (2m - 2\sqrt{m^2 + 1})}$$

This simplifies to:

$$m_{MN} = \frac{(2m + 2\sqrt{m^2 + 1} - (2m - 2\sqrt{m^2 + 1}))}{4\sqrt{m^2 + 1}} = \frac{4\sqrt{m^2 + 1}}{4\sqrt{m^2 + 1}} = 1.$$

Consequently, the slope $m_{FA}$ of line $FA$ must be $-1$ (since $m_{FA} \cdot m_{MN} = -1$):

$$y - 1 = -1(x - 0) \quad \Rightarrow \quad y = -x + 1.$$

We must find the intersection of this line $y = -x + 1$ with the parabola $x^2 = 4y$.

Substituting $y = -x + 1$ into the parabola’s equation gives:

$$x^2 = 4(-x + 1).$$

Rearranging leads to:

$$x^2 + 4x - 4 = 0.$$

Applying the quadratic formula yields:

$$x = \frac{-4 \pm \sqrt{4^2 + 4 \cdot 4}}{2} = \frac{-4 \pm \sqrt{16 + 16}}{2} = \frac{-4 \pm \sqrt{32}}{2} = -2 \pm 2\sqrt{2}.$$

Taking the positive root in the first quadrant:

$$x_A = -2 + 2\sqrt{2}.$$

Then utilizing $y = -x + 1$:

$$y_A = -(-2 + 2\sqrt{2}) + 1 = 2 - 2\sqrt{2} + 1 = 3 - 2\sqrt{2}.$$

Thus, the coordinates of point $A$ are

$$\boxed{(-2 + 2\sqrt{2}, 3 - 2\sqrt{2})}.$$