[Member (365WT)]answers [Chinese ]  Time :20191011  area
S = π (pi) × a × b (where a, b are the long semiaxes of the ellipse and the length of the short semiaxes, respectively).
Or S = π (pi) = A × B / 4 (where A, B are the long axis of the ellipse and the length of the short axis, respectively).
perimeter
There is no formula for the elliptical perimeter, and there are integral or infinite expansions.
The exact calculation of the elliptical perimeter (L) uses the sum of the integral or infinite series. Such as
L = ∫[0,π/2]4a * sqrt(1(e*cost)2)dt≈2π√((a2 b2)/2) [elliptical approximate perimeter], where a is the ellipse long semiaxis, e is the eccentricity
Ellipse eccentricity is defined as the ratio of the focal length of the ellipse to the long axis (range: greater than 0 is less than 1)
The line rule of the ellipse x=±a^2/c
Eccentricity e=c/a (0<e<1) because 2a>2c. The larger the eccentricity, the flatter the ellipse; the smaller the eccentricity, the closer the ellipse is to a circle.
The focal length of the ellipse: the distance between the focus of the ellipse and its corresponding guideline (such as the focus (c, 0) and the guideline x = a^2/c) is b^2/c
Cone radius
The focus is on the xaxis: _PF1_=a ex _PF2_=aex(F1, F2 are left and right focus respectively)
The radius of the ellipse over the right focus r=aex
The radius of the left focus r=a ex
The focus is on the yaxis: _PF1_=aey _PF2_=a ey(F1, F2 are the up and down focus respectively)
The path of the ellipse: the distance between the line perpendicular to the xaxis (or yaxis) of the overfocus and the two intersections A, B of the ellipse, ie _AB_=2*b^2/a
Slope
The tangent slope of the point (x, y) above x^2/a^2 y^2/b^2=1 over the ellipse is (b^2)X/(a^2)y
Triangle area If there is a triangle, two vertices are at the two focal points of the ellipse, and the third vertex is on the ellipse.
Then, if F1PF2 = θ, then S = (b^2) tan (θ/2).
Curvature
K=ab/[(b^2a^2)(cosθ)^2 a^2]^(3/2) 
