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The focus of a parabola is [tex]$(0, -2)$[/tex]. The directrix is the line [tex]$y = 0$[/tex]. What is the equation of the parabola in vertex form?

In the equation [tex]$y = \frac{1}{4p}(x - k)^2 + h$[/tex], the value of [tex][tex]$p$[/tex][/tex] is [tex]$\square$[/tex].

The vertex of the parabola is the point [tex]$(\square, \square)$[/tex].

The equation of this parabola in vertex form is [tex]$y = \square x^2 - 1$[/tex].


Sagot :

Let's work through the problem of finding the equation of a parabola in vertex form, given the focus and the directrix.

The focus of the parabola is [tex]\((0, -2)\)[/tex] and the directrix is the line [tex]\(y = 0\)[/tex].

1. Determine the Vertex:
The vertex of the parabola is midway between the focus and directrix. The focus is at [tex]\(y = -2\)[/tex] and the directrix is at [tex]\(y = 0\)[/tex]. The midpoint between [tex]\(-2\)[/tex] and [tex]\(0\)[/tex] is:
[tex]\[ \text{Vertex} = (0, -1) \][/tex]

2. Calculate [tex]\(p\)[/tex]:
The distance [tex]\(p\)[/tex] is the distance from the vertex to the focus (or to the directrix). In this case, [tex]\(p\)[/tex] is the absolute distance between [tex]\(-1\)[/tex] and [tex]\(-2\)[/tex]:
[tex]\[ p = 1 \][/tex]

3. Write the Equation:
The standard vertex form of a parabola with vertex [tex]\((h, k)\)[/tex] and focus-distance [tex]\(p\)[/tex] is:
[tex]\[ y = \frac{1}{4p}(x - h)^2 + k \][/tex]
Here, [tex]\(h = 0\)[/tex], [tex]\(k = -1\)[/tex], and [tex]\(p = 1\)[/tex]. Plugging in these values, we get:
[tex]\[ y = \frac{1}{4(1)}(x - 0)^2 - 1 = \frac{1}{4}x^2 - 1 \][/tex]

So, the values are:
- [tex]\( p = 1 \)[/tex]
- The vertex is [tex]\((0, -1)\)[/tex]
- The equation of the parabola in vertex form is [tex]\( y = \frac{1}{4}x^2 - 1 \)[/tex]

Now, filling in the boxes:
1. The value of [tex]\(p\)[/tex] is [tex]\(\boxed{1}\)[/tex].
2. The vertex of the parabola is the point [tex]\((\boxed{0}\, \boxed{-1})\)[/tex].
3. The equation of the parabola in vertex form is [tex]\( y = \boxed{\frac{1}{4}} x^2 - 1 \)[/tex].