Question

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

a. y = -1/4 x^2 + 2x - 6
b. y = -1/4 x^2 + 3x -10
c. y = 1/4 x^2 + 3x + 8
d. y = 1/4 x^2 + x + 4

Answer 1

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

1. **Identify the focus and directrix**:
- The focus is at \((6, -2)\).
- The directrix is the line \(y = 0\).

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

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

4. **Set the distances equal**:
\[
\sqrt{(x - 6)^2 + (y + 2)^2} = |y|
\]

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

6. **Expand the left-hand side**:
\[
(x - 6)^2 + (y + 2)^2 = (x - 6)^2 + (y^2 + 4y + 4)
\]
This becomes:
\[
(x - 6)^2 + y^2 + 4y + 4 = y^2
\]

7. **Subtract \(y^2\) from both sides**:
\[
(x - 6)^2 + 4y + 4 = 0
\]
Simplifying gives:
\[
(x - 6)^2 + 4y = -4
\]

8. **Isolate \(y\)**:
\[
4y = - (x - 6)^2 - 4
\]
\[
y = -\frac{1}{4}(x - 6)^2 - 1
\]

9. **Expand it**:
\[
y = -\frac{1}{4}(x^2 - 12x + 36) - 1
\]
\[
y = -\frac{1}{4}x^2 + 3x - 9 - 1
\]
\[
y = -\frac{1}{4}x^2 + 3x - 10
\]

Thus, the derived equation of the parabola is:

\[
y = -\frac{1}{4}x^2 + 3x - 10
\]

This matches option **b**.

**Final answer:**
**b.** \(y = -\frac{1}{4} x^2 + 3x -10\)