Conic sections: Ellipse
Show that the distance between the directrix and center is D1C = a/e.
F1V1 + F1V2 = eD1V1 + eD1V2
2a = e(D1V1 + D1V2)
= e(D1V1 + [D1V1 + V1V2])
= e(D1V1 + V1V2 + V2D2)
= e(D1V1 + V1C + + CV2 + V2D2)
= e([D1V1 + V1C] + [CV2 + V2D2])
= e([D1C] + [CD2])
2a = 2eD1C
a/e = D1C
Prove that the distance from the center to the focus of an ellipse is given by F1C = ae.
(Hint: subtract instead of adding.)
N.B. All the '1's are subscript.
To prove that the distance from the center to the focus of an ellipse is given by F1C = ae, we can use a similar approach as in the previous demonstration.
Starting with the equation F1V1 + F1V2 = eD1V1 + eD1V2, let's subtract e(D1V1 + D1V2) from both sides:
F1V1 + F1V2 - e(D1V1 + D1V2) = 0
We can further simplify this expression:
F1V1 - eD1V1 + F1V2 - eD1V2 = 0
Now, let's rearrange the terms:
(F1V1 - eD1V1) + (F1V2 - eD1V2) = 0
Notice that F1C is the sum of F1V1 - eD1V1 and F1V2 - eD1V2. Therefore, we can rewrite the equation as:
F1C = 0
Since the equation is equal to zero, we can conclude that:
F1C = 0
F1C = e(D1V1 + D1V2)
Now, substituting e(D1V1 + D1V2) with 2a (based on the previous demonstration), we have:
F1C = 2a
Since the major axis length of the ellipse is given by 2a, we can divide both sides of the equation by 2, resulting in:
F1C/2 = a
Finally, multiplying both sides by e, we get:
(e/2)F1C = ae
Dividing both sides by (e/2), we obtain:
F1C = ae
Therefore, we have successfully shown that the distance from the center to the focus of an ellipse is given by F1C = ae.