How to find the shortest distance between focus and ellipse? How to judge which point on the ellipse has the shortest distance to the focus? How to find the shortest distance from the point to the focus?

How to find the shortest distance between focus and ellipse? How to judge which point on the ellipse has the shortest distance to the focus? How to find the shortest distance from the point to the focus?

The vertex of the corresponding focus is closest to the focus (such as the right point nearest to the right focus, the right point nearest to the right focus), and now we can prove that the right vertex is closest to the right focus: let a point on the ellipse (ACOS θ, bsin θ) and the right focus (C, 0), then d (C, 0, C, 0), then d = (ACOS θ - 178; = (ACOS θ - C) = (ACOS θ - 0) ((bsin θ - 0) - 0) = (bsin θ - 0) = (bsin θ - 0) - 0) ((the right vertex is closest to the right focus: let a point on the ellipse (ACOS (a (0, bsin θ, bsin θ, bsin θ, bsin θ, bsin θ, bsin θ, bsin θ, bsin θ, bsin θ, bsin θ, bsin θ, bsin θ, bsin θ: let (Chen Chen 178; (1-cos \# 178; (1-cos \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\whend is the smallest and is (A-C) & 178;, so D is the smallest a-c

What is the shortest distance from the focus to the point on the ellipse? Is it a-c? Why

Yes,
The coordinates of the points on the ellipse are (ACOS @, bsin @)
It can be calculated by the formula of distance between points

Why is the shortest distance from the moving point P on the ellipse to the right focus a-c?

Let the left focus be (f1,0) and the right focus be (f1,0). According to the elliptic property, we know that pf2-pf1 = 2A, a is invariable. When Pf1 is maximum, PF2 is minimum. When P point moves to the far right, Pf1 is maximum a + C. at this time, PF2 = 2a-pf1 = a-c