Thank you for sharing the image. Let's go over the questions one by one and start solving them:

1) Given:

• Point $P(4, 6)$ is a point on a parabola with focus at $F(0, 12)$ and the directrix is parallel to the x-axis.

(a) Find the equation of the parabola, its vertex, and the distance from point $P$ to the directrix.

• The general equation of a parabola is $y = a(x - h)^2 + k$, where $(h, k)$ is the vertex of the parabola, and the distance from any point on the parabola to the focus is equal to the distance from the point to the directrix.

To determine the equation, we'll use the given focus and directrix information.

(b) Determine the distance from point $P$ to the directrix.

We will calculate the distance between the point $P(4, 6)$ and the directrix based on the equation of the parabola.

2) Given the equation:

$x^2 + 2y^2 + 4xy + 8x - 169 + F = 0$

(a) Find all values of $F$ such that the graph of the equation represents an ellipse.

To determine the values of $F$ for which this equation represents an ellipse, we need to analyze the discriminant of the conic equation $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$, where the discriminant is $B^2 - 4AC$. For an ellipse, this discriminant must be negative.

(b) Find the values of $F$ that represent a point.

For this, we'll evaluate when the discriminant and other terms lead to a degenerate form (a point).

(c) Find values of $F$ where the graph consists of no points at all.

This can happen when the discriminant is positive, leading to an equation with no real solutions.

3) Find an equation of a hyperbola:

• Major axis parallel to the x-axis, vertices at $(6, 1)$, and a focus at $(6, -1)$.

We'll derive the equation of a hyperbola with these conditions.

4) Find the equation of an ellipse:

• The foci are at $(2, 0)$ and $(2, 6)$, and the length of the major axis is 5.

The general equation for an ellipse is $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$, where $(h, k)$ is the center and $a$ and $b$ are the lengths of the axes. We will use the given conditions to find the equation.

5) Given the equation of a parabola:

$x^2 = 36y$

Find the area of the triangle formed by the vertex and the endpoints of the latus rectum.

We will calculate the endpoints of the latus rectum and then use the formula for the area of a triangle to find the area.

I will start working on these problems. Do you want me to solve them one at a time, or would you prefer I tackle multiple at once?