Some Proofs
d(F1, P) + d(F2, P) =2a
{the length of the distances from the two foci to the point added together equals the distance of the major axis}
a = the distance of half of the major axis (from the origin to a vertex)
b = the distance of half of the minor axis (from the origin to a co-vertex)
c = the distance from the origin to a focus
Proof:
For further assistance, you can refer to the following video:
http://www.youtube.com/watch?v=OUZ7hUzzYMg
{the length of the distances from the two foci to the point added together equals the distance of the major axis}
a = the distance of half of the major axis (from the origin to a vertex)
b = the distance of half of the minor axis (from the origin to a co-vertex)
c = the distance from the origin to a focus
Proof:
- Let point P be (c, 0)
- d(F1, P) = a + c
- d(F2, P) = a - c
- d(F1, P) + d(F2, P) = a + c + a - c = 2a
For further assistance, you can refer to the following video:
http://www.youtube.com/watch?v=OUZ7hUzzYMg
Let point P be (0, b).
Note that the values of a, b, and c are the same as in the previous equation. In the previous equation, it was proved that 2a = d(F1, P) + d(F2, P), so the hypotenuse of the triangle with legs b and c would be equal to the value of a.
By the Pythagorean theorem, the value of a would equal the square root of b^2 + c^2. Thus, you can draw the formulas below:
Note that the values of a, b, and c are the same as in the previous equation. In the previous equation, it was proved that 2a = d(F1, P) + d(F2, P), so the hypotenuse of the triangle with legs b and c would be equal to the value of a.
By the Pythagorean theorem, the value of a would equal the square root of b^2 + c^2. Thus, you can draw the formulas below:
Working with ellipses with a center at the origin
Here, we are going to prove the standard formula of an ellipse with center (0,0) and its major axis on the x-axis.
Using the picture on the right:
Use the distance formula to add together the distances from the foci to P and note that the sum is 2a. √(〖(x+c)〗^2+y^2 )+√(〖(x-c)〗^2+y^2 )=2a
... simplify until you get to:
a^2 (a^2-c^2 )= x^2 (a^2-c^2 )+a^2 y^2
Remember that b^2=a^2-c^2 and substitute.
a^2 (b^2)=x^2 (b^2 )+a^2 y^2
Divide both sides by (a^2)(b^2) to get the standard form of an ellipse with its major axis on the x-axis.
Using the picture on the right:
Use the distance formula to add together the distances from the foci to P and note that the sum is 2a. √(〖(x+c)〗^2+y^2 )+√(〖(x-c)〗^2+y^2 )=2a
... simplify until you get to:
a^2 (a^2-c^2 )= x^2 (a^2-c^2 )+a^2 y^2
Remember that b^2=a^2-c^2 and substitute.
a^2 (b^2)=x^2 (b^2 )+a^2 y^2
Divide both sides by (a^2)(b^2) to get the standard form of an ellipse with its major axis on the x-axis.
What happens when the major axis is on the y-axis?
All the same principles apply. In the diagram to the right, the major axis is now the y-axis and the minor axis is the axis. The foci are on the y-axis.
The equation for an ellipse with its center at the origin and major axis as the y-axis is:
x^2/b^2 +y^2/a^2 =1
All the same principles apply. In the diagram to the right, the major axis is now the y-axis and the minor axis is the axis. The foci are on the y-axis.
The equation for an ellipse with its center at the origin and major axis as the y-axis is:
x^2/b^2 +y^2/a^2 =1
Working with ellipses with a center at (h, k)
Remember the transformation properties:
Since ellipses with centers at (h, k) are just translations of ellipses with centers at the origin, you can pretty much use the same equations for standard form; you just need to add your (h, k) values.
- To transform to the left, you subtract from the x value.
- To transform to the right, you add to the x value.
- To transform up, you subtract from the y value.
- To transform down, you add from the y value.
Since ellipses with centers at (h, k) are just translations of ellipses with centers at the origin, you can pretty much use the same equations for standard form; you just need to add your (h, k) values.