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.

invalid Aug 29, 2024

#1**0 **

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})}.

\]

gnistory Aug 29, 2024