How to derive the equation of a parabola with a focus at (6, 2) and a directrix of y = 1?

To derive the equation of the parabola with a given focus and directrix, we need to follow the definition of a parabola: it is the set of all points that are equidistant from the focus and the directrix.

Here, the focus is at the point (6, 2), and the directrix is the line y = 1. The distance from any point (x, y) on the parabola to the focus (6, 2) must equal the perpendicular distance from (x, y) to the directrix, which is the line y = 1.

1. **Distance to the focus**:

The distance from a point (x, y) to the focus (6, 2) is given by:

Dfocus = √((x – 6)² + (y – 2)²)

2. **Distance to the directrix**:

The distance from the point (x, y) to the directrix (y = 1) is simply the vertical distance:

Ddirectrix = |y – 1|

3. **Setting the distances equal**:

Since both distances are equal for points on the parabola, we can set up the equation:

√((x – 6)² + (y – 2)²) = |y – 1|

4. **Squaring both sides** to eliminate the square root:

(x – 6)² + (y – 2)² = (y – 1)²

5. **Expanding both sides**:

(x – 6)² + (y² – 4y + 4) = (y² – 2y + 1)

6. **Simplifying**:

(x – 6)² + y² – 4y + 4 = y² – 2y + 1

Now, if we subtract y² from both sides:

(x – 6)² – 4y + 4 = -2y + 1

7. **Bringing all terms involving y to one side:**

(x – 6)² + 2y + 3 = 0

Finally, isolate y:

2y = -(x – 6)² – 3

y = -½(x – 6)² – ¾

Thus, the equation of the parabola with a focus at (6, 2) and a directrix of y = 1 is:

y = -½(x – 6)² – ¾

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