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# parabola problem

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The focus of the parabola x^2=4y is F=(0,1). A line passing through F intersects the parabola at M and N. The line passing through F that is perpendicular to line MN intersects the parabola at a point A in the first quadrant. If angle MAN=90 degrees then enter the coordinates of A.

Aug 29, 2024

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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:

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

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

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

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

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

Aug 29, 2024