Chord length problem of elliptic focus Derivation of focus chord formula

Chord length problem of elliptic focus Derivation of focus chord formula

For example, the parabola is 2P / (the square of sina), and the polar coordinate equation of ellipse is: R (a) = EP / (1-ecosa), where e is the eccentricity of ellipse, P is the distance from the focus to the corresponding collimator, and a is

How to prove that the maximum distance from a point on an ellipse to the focus is a + C and the minimum distance is a-c?

It is proved that: let the elliptic equation: (X & sup2 / A & sup2;) + (Y & sup2 / B & sup2;) = 1. (a > b > 0). Point P (acost, bsint), (t ∈ R), from the symmetry, we can set the focus as f (C, 0). Then | PF | & sup2; = (acost-c) & sup2; + (bsint) & sup2; = (a-ccost) & sup2; = = = = = = > | PF | = a-ccost. | PF | max = a + C, | PF | min = a-c

It is proved by coordinate method that the points on the ellipse with the maximum and minimum distance from the two focal points are exactly the two ends of the major axis of the ellipse

Take the ellipse with focus on X axis as an example. Let the equation be X & # 178 / A & # 178; + Y & # 178 / B & # 178; = 1 (a > b > 0), let P (x, y) be the ellipse, be any point, and F1 (- C, 0) be the left focus. Since X & # 178 / A & # 178; + Y & # 178 / B & # 178; = 1, we can make x = A & # 8226; Cos θ, y = B & # 8226; sin θ, θ ∈ [0,2