I actually have 5 questions on the same thing, just different setups. But how would I go about solving them?.-.

1. Derive the equation of the parabola with a focus at (-5, - 5) and a directrix of y = 7

2. Derive the equation of the parabola with a focus at (6, 2) and a directrix of y = 1

3. Derive the equation of the parabola with a focus at (-5, 5) and a directrix of y = - 1.

4. Derive the equation of the parabola with a focus at (0, - 4) and a directrix of y = 4

5. Derive the equation of the parabola with a focus at (-2, 4) and a directrix of y = 6. Put the equation in standard form.

Question

Mathematics

Jada Kline

21 August, 08:50

I actually have 5 questions on the same thing, just different setups. But how would I go about solving them?.-.

1. Derive the equation of the parabola with a focus at (-5, - 5) and a directrix of y = 7

2. Derive the equation of the parabola with a focus at (6, 2) and a directrix of y = 1

3. Derive the equation of the parabola with a focus at (-5, 5) and a directrix of y = - 1.

4. Derive the equation of the parabola with a focus at (0, - 4) and a directrix of y = 4

5. Derive the equation of the parabola with a focus at (-2, 4) and a directrix of y = 6. Put the equation in standard form.

+5

Answers (1)

Know the Answer?

Raul Li · Answer 1

1) 2p = - 2.

4p [ y - k ] = [ x - h) ]² - - - > - 4 [ y + 5 ] = [ x + 5 ]²

2) 4p * (y - k) = (x - h) ^2

(h, k) is the vertex

The vertex is halfway between the focus and the directrix (when they're at their closest)

p is that distance

2 - 1 = 1

4p = 1

p = 1/4

(1/4) * (y - k) = (x - h) ^2

y - k = 4 * (x - h) ^2

The vertex is at (6, 3/2), since that's midway between (6, 1) and (6, 2)

y - 3/2 = 4 * (x - 6) ^2

y = (3/2) + 4 * (x - 6) ^2

4) f (x) = (-1/16) * (x²)

5) f (x) = - 1/4 x2 - x + 5