Find a parametric description for the ellipse having the focus F=(0,-4), corresponding directrix y=3, and eccentricity e=3/4.
To find a parametric description for the ellipse, we need to determine the equation of the ellipse first.
Since the focus is F = (0, -4) and the directrix is y = 3, we can conclude that the ellipse is vertically oriented.
The distance between the focus and the directrix is given by the formula:
d = |2ae|
Where "a" is the semi-major axis and "e" is the eccentricity.
In this case, we know that e = 3/4, so we can solve for "a":
d = |2(a)(3/4)|
|3 - (-4)| = 6a/4
7 = 6a/4
a = 7 * 4 / 6 = 4.67
Now that we have "a," we can find "b" by using the formula:
b = sqrt(a^2 - c^2)
Where "c" is the distance between the center and the focus (in this case, c = 4).
b = sqrt(4.67^2 - 4^2)
b = sqrt(21.77 - 16)
b = sqrt(5.77)
b ≈ 2.40
Now that we have "a" and "b", we can write the equation of the ellipse in standard form:
(x - h)^2 / a^2 + (y - k)^2 / b^2 = 1
In this case, the center (h, k) is (0, (-4 + 3) / 2) = (0, -0.5).
(x - 0)^2 / (4.67^2) + (y - (-0.5))^2 / (2.40^2) = 1
Simplifying, we get:
x^2 / 21.77 + (y + 0.5)^2 / 5.77 = 1
Now, to parametrize the equation, we can substitute:
x = a * cos(t)
y = b * sin(t)
Where "t" is the parameter.
So the parametric equations for the ellipse with the given focus, directrix, and eccentricity are:
x = 4.67 * cos(t)
y = 2.40 * sin(t) - 0.5