Isaiah's Math Lessons : Ellipse
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- The locus of a point that moves such that the sum of its distances from two fixed points called the foci is constant. The constant sum is equal to 2a, which is the length of the major axis.
Ax2 + Cy2 + Dx + Ey + F = 0
KEY FORMULAS AND SHORTCUTS:
2a = Length of major axis
2b = Length of minor axis
2c = distance between the foci
2b2/a = Length of Latus Rectum
d = a/e = distance from center to directrix
e = c/a = eccentricity of ellipse (between 0 and 1 for ellipse)
a2 = b2 + c2 where c is the distance from center to focus
Area = pi*a*b
Examples and Detailed Solutions:
1) What is the general equation of an ellipse with Center at (2, 0), focus at (5, 0) and b = 4?
We can solve first for c since the center and the focus is given. We just need to get the distance.
Thus c = 3.
a2 = b2 + c2
a2 = 42 + 32
a = 5
If we graph this, we will notice that the axis is horizontal. (Since there was no change on y in focus and center)
2. Find the equation of the locus of a point which moves so that the sum of its distances from the points (2, 1) and (8, 1) is 10.
Based from the definition, the two fixed points are the foci. Thus, the distance between them is 2c. And the sum is what we called as the major axis or 2a.
Distance from (2, 1) and (8, 1) is 6
2c = 6
c = 3
2a = 10
a = 5
a2 = b2 + c2
52 = b2 + 32
b = 4
To get the Center, get the midpoint of the foci. Midpoint = ((x1+x2)/2, (y1+y2)/2) = ((2+8 )/2, (1+1)/2)
C is @ (5, 1).
Since the y-value of the foci didn’t change, the axis is horizontal.
1. What is the eccentricity of the ellipse 4x2 + 3y2 – 48 = 0?
2. What is the general equation of the ellipse with foci at (2, 1) and (2, -1) and a = 2?
3. Is the point (1, 1) inside or outside the ellipse 4x2 + 9y2 – 24x – 36y + 36 = 0?
2. 4x2 + 3y2 – 16x + 4 = 0