If the focal length of the ellipse is equal to the distance between one end of the major axis and one end of the minor axis, then the eccentricity of the ellipse is 0

If the focal length of the ellipse is equal to the distance between one end of the major axis and one end of the minor axis, then the eccentricity of the ellipse is 0

√(a^2 + b^2) = 2c ,∴a^2 + b^2 = 4c^2 ,∴2a^2 - c^2 = 4c^2 ,∴e^2 = c^2/a^2 = 2/5
∴e = (√10)/5

If the focal length of the ellipse is equal to the distance between one end point of the major axis and one end point of the minor axis, find E

2C = a ^ 2 + B ^ 2 under root sign
4c^2=a^2+b^2
E = root 10 / 5

If the focal length of an ellipse is equal to half of the distance between two collimators, then the eccentricity of the ellipse is zero

The elliptic quasilinear equation is x = (+ / -) a ^ 2 / C
The distance between the two guide lines is 2A ^ 2 / C
The focal length is 2C
(2a^2/c)/2=2c
a^2/c=2c
a^2/c^2=2
c^2/a^2=1/2
e=c/a
The solution is e = root 2 / 2

The left vertex of the ellipse of equation x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 is a, and the left and right focuses are F1, F2 and D respectively. If 2df1 vector = Da vector
If 2df1 vector = Da vector + df2 vector, the eccentricity of the ellipse is

Because: 2df1 vector = Da vector + df2 vector
AF1=F1F2=2c
AF2=4C
AF1+AF2=2a
2c+4c=2a
6c=2a
e=c/a=1/3

Ellipse x ^ 2 / 3 + y ^ 2 = 1, line y = x intersects ellipse at two points a and B, C is the right vertex of ellipse, vector OA · vector OC = 3 / 2, if two points E and F on ellipse are oriented
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Did you come to talk to me just now? I'm still not sure. I've read all the answers on the Internet
If two points E and F on the ellipse make vector OE + vector of = λ vector OA, λ∈ (0,2), the maximum area of triangle OEF is obtained

2 radical 3

In the following four propositions, the correct one is ()
A. If f '(x) is continuous in (0,1), then f (x) is bounded in (0,1). B. if f (x) is continuous in (0,1), then f (x) is bounded in (0,1). C. if f' (x) is bounded in (0,1), then f (x) is bounded in (0,1). D. if f (x) is bounded in (0,1), then f '(x) is bounded in (0,1)

Solution (1) Let f (x) = 1 x, then f (x) and f ′ (x) = − 1 x 2 are continuous in (0,1), but f (x) is unbounded in (0,1), excluding (a), (b); & nbsp; & nbsp; (2) Let f (x) = x, then f (x) is bounded in (0,1), but f ′ (x) = 12 x is unbounded in (0,1), excluding (d)

A higher number, about Rolle's theorem, or Lagrange's mean value theorem
Let f (x) be continuous on [0, π / 4], differentiable on (0, π / 4), and f (π / 4) = 0. It is proved that there is a point C ∈ (0, π / 4) such that 2F (c) + sin2c × f '(c) = 0

Let f (x) = f (x) * TaNx, 0

If the function f is differentiable and f (0) = 0, | f '(x) | ＜ 0, we should use Rolle's theorem or Lagrange's mean value theorem to prove higher numbers
Proof of Higher Mathematics
Use Rolle's theorem or Lagrange's mean value theorem
If the function f is differentiable and f (0) = 0, | f '(x) | 1, it is proved that when x is not equal to 0, | f (x) | X|

Using Lagrange or roll theorem to prove
It is proved that the cube-3x + C = 0 of X cannot have two different real roots in the closed interval 0 to 1

∵x^3-3x+c=0
∴f(x)’=3x^2-3=3(x+1)(x-1)
∴f(x)’’=6x
∴x∈（-∞,-1）（-1,0）（0,1）（1,∞）
When x ∈ [0,1],
Monotonic decreasing of F (x)
Therefore, f (x) has only one real root of X

∫∫ ZDS, where ∑ is the part of the paraboloid z = 2 - (X & # 178; + Y & # 178;) above xoy
∫∫ ZDS, where ∑ is a paraboloid, z = 2 - (X & # 178; + Y & # 178;) above xoy ∑ I can't keep up with the teacher's progress and understand the meaning of the topic,

az/ax=-2x
az/ay=-2y
dS=√1+4x²＋4y² dxdy
therefore
The original formula = ∫ d [2 - (X & # 178; + Y & # 178;)] √ (1 + 4x & # 178; + 4Y & # 178;) DXDY. This is a double integral
Then use polar coordinates to solve