Question

Use the given information to write the equation of the parabola.

Vertex:
`(0,0)`
Focus:
`(0,-(1)/(2))`

Answer 1

x² = -2y

Explanation:

The focus is p away from the vertex, and so is the directrix.

To find the equation of the parabola, we must first determine if the parabola is horizontal or vertical.

• Horizontal parabola [Standard form]: (y – k)² = 4p(x – h)
• Vertical parabola [Standard form]: (x – h)² = 4p(y – k)

If the parabola is vertical, the directrix, and focus will have the same x value but different y value compared to the vertex (h, k). You can also tell if the directrix in in the form y = k – p, and if the focus is in the form (h, k + p).

Likewise, if the parabola is horizontal, the directrix, and focus will have the same y value but different x value compared to the vertex (h,k) . You can also tell if the directrix is in the form x = h – p, and if the focus is in the form (h + p, k).

For this problem, given that the vertex is at the origin (0,0), and that the focus is at the point (0, -½).

You can tell that the x value is the same for the vertex, and focus so this must be a vertical parabola. Because this is a vertical parabola, we can use the form mentioned as earlier (x – h)² = 4p(y – k).

If h = 0, and k = 0, the p value must be the difference between the k of the vertex, and the k of the focus: -½ - 0 → -½.

Now we can just plug in our known information to derive the equation!

h = 0, k = 0, p = -½ → (x - h)² = 4p(y - k) →

(x - 0)² = 4(-½)(y - 0) → x² = -2(y - 0) →

x² = -2y.

Also p = 1/4a, if you are wondering.

So because this is a vertical parabola, x² = -2y is generally the same as y = -1/2x² in standard quadratic form. I just like to think of the horizontal parabola as an inverse quadratic because it is like reflecting over the line y = x.

Answer 2

`\boxed {\boxed {\sf x^2=-2y}}`

Explanation:

First, plot the two points. We can see the focus is below the vertex. The result will be a vertical parabola. Since the parabola must contain the focus, it will have to open downward.

From this information, we know that the following equation must be used:

`(x-h)^2=-4p(y-k)`

where (h,k) is the vertex and p is the distance from the focus to the vertex.

Next, find p, the distance between the focus and vertex. We can use the distance formula.

`d=√((x_2-x_1)^2+(y_2-y_1)^2)`

The vertex is (0,0) and the focus is (0, -1/2). Therefore:

`x_1=0\\y_1=0\\x_2=0\\y_2= -1/2`

`d=√((0-0)^2+(0--1/2)^2)`

`d=√((0)^2+(0+1/2)^2)`

`d=√(0+(1/2)^2)`

`d=√(0+1/4)`

`d=√(1/4)`

`d=1/2`

Now we know the distance and vertex. We can plug the values into the equation.

d=1/2 and (h,k)= (0,0)

`(x-0)^2=-4(1/2)(y-0)`

`(x)^2=-2(y)`

`x^2=-2y`

The equation of the parabola is x²= -2y