1)Find the equation of a parabola whose vertex is (0,0) and directrix is the line y=3

1. Find the equation of a parabola whose vertex is (0,0) and directrix is the line y=3

Since the directrix line is y = 3, we know that this is a parabola, which opens down. Vertex is (00), so we also know that this curve is symmetric with respect to the y axis. So the equation should be of the form of x^2 = 4a y with y = a being the focus ( Note that for each value of y there are two values of x. x = +/- sqrt (4ay), this is another trick to right down the general formula). Lets find the value of a. y = +3 is the directrix. Hence y = -3 is the focus. Hence a = -3.

Equation is x^2 = 4a y = 4(-3) y = -12y

x^2 = -12y

2. Find the vertex, focus, and directrix of (x-2)^2=12(y+1). Find the latus rectum and graph the parabola, making sure that all points and axis are labeled.

(x-2)^2=12(y+1)

It is easy to work with a parabola of the form of x^2 = 4a y. We can make the given equation look like this by shifting the coordinates. Lets say X = x -2 and Y = y +1.

X^2=12 Y = 4 (3) Y

This is a familiar form. Vertex is given by X=0 and Y = 0. That is, x -2 =0 and y +1. That is, x =2 and y = -1
Hence Vertex = (2,-1).

X^2 = 4 (3) Y

This is a parabola, which opens up. Focus is on the Y axis. Focus is, (X,Y) = (0,+3). That is, (x-2,y+1) = (0,+3). è (x,y) = (2, 2)

Directrix is Y = -3. i.e. y+1 = -3. i.e. y = -4

Latus rectum = 4a = 12

3. Find the equation of the ellipse whose center is the origin and has a vertex at (0,5) and a focus at (0,3).

Since the vertex is in the y axis, this is an ellipse with its major axis vertical. Since the vertex is (0,5), we know that the major axis pass through the origin, which means major axis is y axis.

General form of this kind of ellipse is x^2/b^2 + y^2/a^2 = 1 with a > b.

Vertex (0,5) indicates that a = 5.

Focal length = sqrt (a^2 - b^2) = 3

(5^2 - b^2) = 9

solve for b: b = 4

Equation is : x^2/16 + y^2/25 = 1.

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