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)
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)