Question
erive the equation of a parabola given the focus is at (-4, 6) and the directris is ar y = 8. Fill in the missing values of the equation in standard form. (1 point) y= −14x2 − x + 3
Answers
Answered by
GPT-4o mini
To derive the equation of a parabola given the focus and the directrix, we can follow these steps. The focus is \((-4, 6)\) and the directrix is \(y = 8\).
1. **Identify the vertex**: The vertex of the parabola lies halfway between the focus and the directrix.
- The y-coordinate of the vertex is halfway between \(6\) (focus) and \(8\) (directrix):
\[
y_{vertex} = \frac{6 + 8}{2} = 7
\]
- The x-coordinate of the vertex is the same as the focus since it is a vertical parabola:
\[
x_{vertex} = -4
\]
- Therefore, the vertex is at \((-4, 7)\).
2. **Calculate the distance \(p\)**: The distance \(p\) from the vertex to the focus (or the vertex to the directrix) is:
\[
p = 6 - 7 = -1
\]
(Note: It is negative since the focus is below the directrix.)
3. **Write the standard form of the parabola**: For a parabola that opens downwards, the standard form is:
\[
(x - h)^2 = -4p(y - k)
\]
where \((h, k)\) is the vertex.
Here, \(h = -4\), \(k = 7\), and \(4p = 4(-1) = -4\).
4. **Plugging in the values**:
\[
(x + 4)^2 = -4(y - 7)
\]
To rewrite in a more standard form:
\[
(x + 4)^2 = -4y + 28
\]
\[
(x + 4)^2 + 4y - 28 = 0
\]
Rearranging, we get:
\[
y = -\frac{1}{4}(x + 4)^2 + 7
\]
5. **Expanding to a general quadratic form**:
To express it in standard quadratic form \(y = ax^2 + bx + c\):
\[
y = -\frac{1}{4}(x^2 + 8x + 16) + 7
\]
\[
= -\frac{1}{4}x^2 - 2x - 4 + 7
\]
\[
= -\frac{1}{4}x^2 - 2x + 3
\]
Thus, the equation of the parabola can be represented (in standard quadratic form) as:
\[
y = -\frac{1}{4}x^2 - 2x + 3
\]
---
Now, if you are interested in filling in the missing values of the equation, in your case it appears to be mismatched with \(y = -14x^2 - x + 3\). However, from our derivation, the correct equation derived from the given focus and directrix is:
\[
y = -\frac{1}{4}x^2 - 2x + 3
\]
Let me know if any further checks or alterations are needed!
1. **Identify the vertex**: The vertex of the parabola lies halfway between the focus and the directrix.
- The y-coordinate of the vertex is halfway between \(6\) (focus) and \(8\) (directrix):
\[
y_{vertex} = \frac{6 + 8}{2} = 7
\]
- The x-coordinate of the vertex is the same as the focus since it is a vertical parabola:
\[
x_{vertex} = -4
\]
- Therefore, the vertex is at \((-4, 7)\).
2. **Calculate the distance \(p\)**: The distance \(p\) from the vertex to the focus (or the vertex to the directrix) is:
\[
p = 6 - 7 = -1
\]
(Note: It is negative since the focus is below the directrix.)
3. **Write the standard form of the parabola**: For a parabola that opens downwards, the standard form is:
\[
(x - h)^2 = -4p(y - k)
\]
where \((h, k)\) is the vertex.
Here, \(h = -4\), \(k = 7\), and \(4p = 4(-1) = -4\).
4. **Plugging in the values**:
\[
(x + 4)^2 = -4(y - 7)
\]
To rewrite in a more standard form:
\[
(x + 4)^2 = -4y + 28
\]
\[
(x + 4)^2 + 4y - 28 = 0
\]
Rearranging, we get:
\[
y = -\frac{1}{4}(x + 4)^2 + 7
\]
5. **Expanding to a general quadratic form**:
To express it in standard quadratic form \(y = ax^2 + bx + c\):
\[
y = -\frac{1}{4}(x^2 + 8x + 16) + 7
\]
\[
= -\frac{1}{4}x^2 - 2x - 4 + 7
\]
\[
= -\frac{1}{4}x^2 - 2x + 3
\]
Thus, the equation of the parabola can be represented (in standard quadratic form) as:
\[
y = -\frac{1}{4}x^2 - 2x + 3
\]
---
Now, if you are interested in filling in the missing values of the equation, in your case it appears to be mismatched with \(y = -14x^2 - x + 3\). However, from our derivation, the correct equation derived from the given focus and directrix is:
\[
y = -\frac{1}{4}x^2 - 2x + 3
\]
Let me know if any further checks or alterations are needed!
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