Question

derive the equation of a parabola with a focus at (6, -2) and a directrix at y = 0

Answer 1

To derive the equation of a parabola given its focus and directrix, we can use the definition of a parabola: it is the set of all points (x, y) that are equidistant from the focus and the directrix.

Here, we have the following:
- Focus: \( F(6, -2) \)
- Directrix: \( y = 0 \)

1. **Distance to the focus**: The distance from a point \( (x, y) \) to the focus \( (6, -2) \) is given by:
\[
d_{\text{focus}} = \sqrt{(x - 6)^2 + (y + 2)^2}
\]

2. **Distance to the directrix**: The distance from a point \( (x, y) \) to the directrix \( y = 0 \) is simply the vertical distance from the point to this line:
\[
d_{\text{directrix}} = |y - 0| = |y|
\]

3. **Set the distances equal**: By the definition of the parabola, we set the distances equal to each other:
\[
\sqrt{(x - 6)^2 + (y + 2)^2} = |y|
\]

4. **Square both sides** to eliminate the square root:
\[
(x - 6)^2 + (y + 2)^2 = y^2
\]

5. **Expand both sides**:
\[
(x - 6)^2 + (y^2 + 4y + 4) = y^2
\]

The \( y^2 \) terms cancel out:
\[
(x - 6)^2 + 4y + 4 = 0
\]

6. **Rearrange the equation**:
\[
(x - 6)^2 + 4 = -4y
\]

Rearranging gives:
\[
4y = -(x - 6)^2 - 4
\]

Which can be simplified further:
\[
y = -\frac{1}{4}(x - 6)^2 - 1
\]

7. **Final equation**: The final equation of the parabola in vertex form is:
\[
y = -\frac{1}{4}(x - 6)^2 - 1
\]

This describes a parabola that opens downwards with a vertex at \( (6, -1) \).