1. Identify the focus and the directrix for 36(y+9) = (x - 5)^2 2. Identify the focus and the directrix for 20(x-8) = (y + 3)^2

Problem 1

Focus: (5, 0)

Directrix:  y = -18

Explanation:

The given equation can be written as 4*9(y-(-9)) = (x-5)^2

Then compare this to the form 4p(y-k) = (x-h)^2

We see that p = 9. This is the focal distance. It is the distance from the vertex to the focus along the axis of symmetry. The vertex here is (h,k) = (5,-9)

We'll start at the vertex (5,-9) and move upward 9 units to get to (5,0) which is where the focus is situated. Why did we move up? Because the original equation can be written into the form y = a(x-h)^2 + k, and it turns out that a = 1/36 in this case, which is a positive value. When 'a' is positive, the focus is above the vertex (to allow the parabola to open upward)

The directrix is the horizontal line perpendicular to the axis of symmetry. We will start at (5,-9) and move 9 units down (opposite direction as before) to arrive at y = -18 as the directrix. Note how the point (5,-18) is on this horizontal line.

Problem 2

Focus:  (13,-3)

Directrix:  x = 3

Explanation:

We'll use a similar idea as in problem 1. However, this time the parabola opens to the right (rather than up) because we are squaring the y term this time.

20(x-8) = (y+3)^2 is the same as 4*5(x-8) = (y-(-3))^2

It is in the form 4p(x-h) = (y-k)^2

vertex = (h,k) = (8,-3)

focal length = p = 5

Start at the vertex and move 5 units to the right to arrive at (13,-3). This is the location of the focus.

Go back to the focus and move 5 units to the left to arrive at (3,-3). Then draw a vertical line through this point to generate the directrix line x = 3

step-by-step explanation:

∆abc has the points a(1, 7), b(-2, 2), and c(4, 2) as its vertices. formed with the point d(1, 2) as its third vertex, then ∆abd is triangle.