Find the focus, directrix, and focal diameter of the parabola.

x2 = 5y

focus (x, y) = 0, 5/4
directrix:y=-5/4 focal diameter=5
Sketch its graph?
my answers seem to be correct but i cannnot sketch the graph correctly.

X^2 = 5y. Y-parabola.

Y = (1/5)x^2
a = 1/5, 4a = 4/5, 1/4a = 5/4.

h = Xv = -b/2a = 0/(2/5) = 0.
k = Yv = (1/5)*0^2 = 0.
V(h,k) = V(0,0).

D(h,Y1), V(h,k), F(h,Y2). Ver. Line.
D(0,Y1), V(0,0), F(0,Y2).

DV = 0-Y1 = 1/4a
0-Y1 = 1/4a = 5/4.
Y1 = -5/4.

VF = Y2-0 = 1/4a.
Y2-0 = 1/4a = 5/4
Y2 = 5/4.

Focal Dia. = 1/a = 1/(1/5) = 5.
A(X1,5/4), F(0,5/4), B(X2,5/4). Hor line

AF = 0-X1 = 5/2
0-X1 = 5/2
X1 = -5/2.

FB = X2-0 = 5/2
X2 = 5/2.

Your answers are correct!

To graph the parabola, select values of
X below and above h(Xv); and calculate
the corresponding value of Y in each
case.

(x,Y). Y = (1/5)x^2.
(-3,9/5).
(-2,4/5).
(-1,1/5).
V(0,0).
(1,1/5).
(2,4/5).
(3,9/5).

Well, it looks like you've got the right answers for the focus and directrix. However, I'm afraid I can't help you with sketching the graph, as I am just a text-based bot and don't have the ability to generate visual images. But hey, don't worry about it! Graphing can be tricky sometimes. Maybe you could use some online graphing tools or ask a friend for assistance. Just remember, even if the sketch doesn't turn out perfectly, it's the effort that counts!

To sketch the graph of the parabola, we can start by finding the vertex, focus, and directrix.

The equation of the parabola is x^2 = 5y, which is in the form (x - h)^2 = 4p(y - k), where (h, k) is the vertex.

Comparing the given equation to the standard form, we have:
h = 0 and 4p = 5.
Since p = 5/4, the vertex is (0, 0).

The focus is located at (0, p). Substituting p = 5/4, the focus is (0, 5/4).

The directrix is the line y = -p. Substituting p = 5/4, the directrix is y = -5/4.

The focal diameter is the distance between the focus and the directrix, which is equal to 2p. Therefore, the focal diameter is 2(5/4) = 5/2.

To sketch the graph, we can plot the vertex, focus, and directrix.

- The vertex is at (0, 0), which is the lowest point on the graph.
- The focus is at (0, 5/4), which is above the vertex.
- The directrix is the horizontal line y = -5/4, which is below the vertex.

Now we can draw the parabola by creating a smooth curve that is symmetrical to the focus and directrix. Since the parabola opens upward, the curve will be U-shaped, with the vertex at the bottom of the "U".

Finally, we can label the vertex, focus, and directrix on the graph.

Please note that without a visual representation, it may be difficult to accurately sketch the graph. It is recommended to use graphing software or tools to visualize the parabola accurately.

To sketch the graph of the parabola with equation x^2 = 5y, you can follow these steps:

1. Identify the vertex: The vertex of the parabola is at the origin (0, 0) since there is no additional term in the equation that shifts the parabola horizontally or vertically.

2. Determine the focus: The focus of the parabola can be found using the formula (h, k + p), where (h, k) represents the vertex and p is the distance from the vertex to the focus. In this case, the vertex is (0, 0) and p = 1/4 of the coefficient in front of y. So, p = 1/4 * 5 = 5/4. Therefore, the focus is at (0, 5/4).

3. Find the directrix: The equation of the directrix can be determined using the formula y = k - p, where k is the y-coordinate of the vertex and p is the same value as above. In this case, the y-coordinate of the vertex is 0, and p = 5/4. So, the equation of the directrix is y = 0 - 5/4 = -5/4.

4. Determine the focal diameter: The focal diameter of a parabola is twice the distance from the vertex to the focus. In this case, the distance from the vertex to the focus is 5/4, so the focal diameter is 2 * (5/4) = 5.

Now, let's sketch the graph of the parabola:

- Mark the vertex at (0, 0).
- Draw a straight line (vertical) passing through the vertex.
- Mark the focus at (0, 5/4) above the vertex.
- Draw a dotted line (horizontal) through the focus.
- Mark the directrix at y = -5/4 below the vertex.
- Draw the parabolic curve starting from the vertex, by slightly curving upwards towards the focus and downwards towards the directrix.

By following these steps, you should be able to sketch the graph of the parabola correctly.