If a parabola has its vertex at the origin and its directrix at y = 200, then what are the coordinates of the parabola's focal point? write your answer as an ordered pair, like this: (2, -14)

Graph x2 = 4y and state the vertex, focus, axis of symmetry, and directrix.This is the same graphing that I've done in the past: y = (1/4)x2. So I'll do the graph as usual: The vertex is obviously at the origin, but I need to "show" this "algebraically" by rearranging the given equation into the conics form:x2 = 4y Copyright © Elizabeth Stapel 2010-2011 All Rights Reserved
(x – 0)2 = 4(y – 0)This rearrangement "shows" that the vertex is at (h, k) = (0, 0). The axis of symmetry is the vertical line right through the vertex: x = 0. (I can always check my graph, if I'm not sure about this.) The focus is "p" units from the vertex. Since the focus is "inside" the parabola and since this is a "right side up" graph, the focus has to be above the vertex.From the conics form of the equation, shown above, I look at what's multiplied on the unsquaredpart and see that 4p = 4, so p = 1. Then the focus is one unit above the vertex, at (0, 1), and the directrix is the horizontal line y = –1, one unit below the vertex.vertex: (0, 0); focus: (0, 1); axis of symmetry: x = 0; directrix: y = –1Graph y2 + 10y + x + 25 = 0, and state the vertex, focus, axis of symmetry, and directrix.Since the y is squared in this equation, rather than the x, then this is a "sideways" parabola. To graph, I'll do my T-chart backwards, picking y-values first and then finding the corresponding x-values for x = –y2 – 10y – 25:To convert the equation into conics form and find the exact vertex, etc, I'll need to convert the equation to perfect-square form. In this case, the squared side is already a perfect square, so:y2 + 10y + 25 = –x 
(y + 5)2 = –1(x – 0)This tells me that 4p = –1, so p = –1/4. Since the parabola opens to the left, then the focus is 1/4 units to the left of the vertex. I can see from the equation above that the vertex is at (h, k) = (0, –5), so then the focus must be at (–1/4, –5). The parabola is sideways, so the axis of symmetry is, too. The directrix, being perpendicular to the axis of symmetry, is then vertical, and is 1/4 units to the right of the vertex. Putting this all together, I get:vertex: (0, –5); focus: (–1/4, –5); axis of symmetry: y = –5; directrix: x = 1/4Find the vertex and focus of y2 + 6y + 12x – 15 = 0The y part is squared, so this is a sideways parabola. I'll get the y stuff by itself on one side of the equation, and then complete the square to convert this to conics form.y2 + 6y – 15 = –12x 
y2 + 6y + 9 – 15 = –12x + 9 
(y + 3)2 – 15 = –12x + 9 
(y + 3)2 = –12x + 9 + 15 = –12x + 24 
(y + 3)2 = –12(x – 2) 
(y – (–3))2 = 4(–3)(x – 2)Then the vertex is at (h, k) = (2, –3) and the value of p is –3. Since y is squared and p is negative, then this is a sideways parabola that opens to the left. This puts the focus 3 units to the left of the vertex.vertex: (2, –3); focus: (–1, –3)