Let F=(0,0) be the focus and the line y=-6 be the directrix. Plot several points P that are half as far from the focus as they are from the directrix. The configuration of all such points is an ellipse. Identify the four points where the ellipse crosses the coordinate axes (two on each axis). Use the distance formula to write an equation for the ellipse.
How I tried solving this: I created two distance equations between a point (x,y) and the focus and the directrix, which I just made into the point (0,-6). When I set those two equations equal to each other while multiplying the equation with the directrix by 1/2, I got x^2+y^2=x+12 which doesn't seem right.
you didn't show your work, but I get for any point (x,y) on the curve:
√(x^2+y^2) = 1/2 (y+6)
x^2+y^2 = 1/4 (y^2+12y+36)
4x^2 + 4y^2 = y^2 + 12y + 36
4x^2 + 3y^2 - 12y = 36
4x^2 + 3(y^2-4y+4) = 36+3*4
4x^2 + 3(y-2)^2 = 48
x^2/12 + (y-2)^2/16 = 1
To plot several points P that are half as far from the focus as they are from the directrix, we can follow these steps:
1. Let's find the coordinates of the focus and the directrix based on the given information. The focus F is located at (0,0), and the directrix is the horizontal line y = -6.
2. Let P(x, y) be a point on the ellipse. The distance between P and the focus F is half the distance between P and the directrix y = -6.
Using the distance formula, we get:
√((x - 0)^2 + (y - 0)^2) = 1/2 * |y - (-6)|
Simplifying this equation:
√(x^2 + y^2) = 1/2 * |y + 6|
Now, let's square both sides of the equation to eliminate the square root:
x^2 + y^2 = 1/4 * (y + 6)^2
Expanding and simplifying:
x^2 + y^2 = 1/4 * (y^2 + 12y + 36)
Multiplying through by 4 to eliminate fractions:
4x^2 + 4y^2 = y^2 + 12y + 36
3x^2 + 3y^2 = 12y + 36
3x^2 + 3y^2 - 12y - 36 = 0
Divide through by 3 to get the equation of the ellipse in standard form:
x^2 + y^2 - 4y - 12 = 0
Now, to find the points where the ellipse crosses the coordinate axes, substitute x = 0 and y = 0 into the equation:
For x = 0:
0^2 + y^2 - 4y - 12 = 0
y^2 - 4y - 12 = 0
Using the quadratic formula, we find the solutions for y:
y = (-(-4) ± √((-4)^2 - 4(1)(-12))) / 2(1)
y = (4 ± √(16 + 48)) / 2
y = (4 ± √64) / 2
y = (4 ± 8) / 2
This gives us two values of y: y = -2 and y = 6
So, the points where the ellipse crosses the y-axis are (0, -2) and (0, 6).
For y = 0:
x^2 + 0^2 - 4(0) - 12 = 0
x^2 - 12 = 0
x^2 = 12
Taking the square root of both sides:
x = ±√12
This gives us two values of x: x = √12 and x = -√12.
So, the points where the ellipse crosses the x-axis are (√12, 0) and (-√12, 0).
Therefore, the four points where the ellipse crosses the coordinate axes are:
(0, -2), (0, 6), (√12, 0), and (-√12, 0).
The equation of the ellipse is x^2 + y^2 - 4y - 12 = 0.