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Vertex, focus and directrix of Parabola and ellipse

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

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.

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).

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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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