Question

A parabola can be drawn given a focus of (-10,1) and a directrix of x = -6. What can be said about the parabola?

Answer 1

Answer:

The characteristics of the parabola are;

1) The vertex of the parabola, (h, k) = (-8, 1)

2) The parabola opens left

3) The axis of symmetry is, y = 1

4) The equation of the parabola is x = -y²/8 + y/4 - 65/8

5) The end points of the latus rectum are (-10, -3), and (-10, 5)

Step-by-step explanation:

The focus of the parabola = (-10, 1)

The directrix of the parabola, x = -6

The directrix; x = h - p

The focus, F(h + p, k)

Therefore, we have;

h + p = -10

h - p = -6

2·h = -16

h = -8

p = -10 - h = -10 - (-8) = -2

p = -2

Therefore, we have;

1) The vertex of the parabola, (h, k) = (-8, 1)

2) Therefore, the parabola opens left

3) The axis of symmetry is the line y = 1

For P < 0, the equation of the parabola is of the form (y - k)² = 4·p·(x - h)

Therefore, the equation of the parabola is (y - 1)² = 4×(-2)×(x - (-8))

(y - 1)² = 4×(-2)×(x - (-8)) = y² - 2·y + 1 = -8·x - 64

∴ y² - 2·y + 1 + 64 = -8·x

x = -y²/8 + y/4 - 65/8

4) The equation of the parabola is x = -y²/8 + y/4 - 65/8

5) The end points of the latus rectum are (h + p, k + 2·p) and (h + p, k - 2·p)

Therefore, the end points of the latus rectum of the given parabola are;

(-8 + (-2), 1 + 2×(-2)) and (-8 + (-2), 1 - 2×(-2))

(-10, -3), and (-10, 5)