Given that the distance from a focal point of an ellipse to the corresponding guide line is equal to the length of the major half axis of the ellipse, then the eccentricity of the ellipse is ()

Given that the distance from a focal point of an ellipse to the corresponding guide line is equal to the length of the major half axis of the ellipse, then the eccentricity of the ellipse is ()

The root of two is 5-1
The distance between a focal point and the corresponding guide line is a ^ 2-c. it is equal to the length of the major half axis, that is, a
So we can get the formula A ^ 2 - C - C = a
Divide both sides by a at the same time, we can get a-c-c-a = 1. Because we need C-A, we can set c-a
If t, we can get 1-T = 1
Given the elliptic equation x ^ 2 / 25 + y ^ 2 / 9 = 1, then the ratio of the distance from a point m on the ellipse to the left focus and the left quasilinear is equal to?
The ratio of the distance from a point m on the ellipse to the left focus and the left quasilinear is the eccentricity
Eccentricity e = C / A
a^2=25 a=5
b^2=9 b=3
c^2=a^2-b^2=25-9=16 c=4
e=c/a=4/5
The ratio of the distance from a point m on the ellipse to the left focus and the left quasilinear is equal to 4 / 5
Given the elliptic equation x ^ 2 / 16 + y ^ 2 / 4 = 1, then the ratio of the distance from a point P on the ellipse to the right focus and the right guide line is equal to?
The denominator of the numerator is 2 / 16 to the power of 4
a²=16,b²=4
Then C & # 178; = A & # 178; - B & # 178; = 12
So the ratio of the distance from point P to the right focus and the right guide line is e = C / a = √ 3 / 2
As shown in the figure, in the plane rectangular coordinate system, the coordinates of vertex B of rhombic oabc are (8,4), then the coordinates of point C are over weighted
Let the side length be x, then x ^ 2 - (8-x) ^ 2 = 4 ^ 2, and the solution is x = 5
C (3,4) area = 20
If the tangent of the curve y = KX + LNX at point (1, K) is parallel to the X axis, then K=______ .
According to the meaning of the question, y '= K + 1 x, ∵ the tangent at point (1, K) is parallel to the X axis, ∵ K + 1 = 0, k = - 1, so the answer is: - 1
Given cos α = 0.68, calculate the values of sin α, Tan α and cot α (the results retain two significant numbers)
sin²α+cos²α=1
So sin α = ± 0.73
tanα=sinα/cosα
cotα=1/tanα
therefore
sinα=0.73,tanα=1.1,cotα=0.93
Or sin α = - 0.73, Tan α = - 1.1, cot α = - 0.93
sinα=0.73,tanα=1.1,cotα=0.93
As shown in the figure, in the plane rectangular coordinate system, the coordinates of vertex B of rhombic oabc are (8,4), then the coordinates of point C are (8,4)______ .
Let AB = x, then OA = x, ad = 8-x. in RT △ abd, AB2 = ad2 + BD2, that is, X2 = (8-x) 2 + 16, the solution is: x = 5, BC = 5, the coordinates of C point are (3,4). So the answer is: (3,4)
In this paper, we discuss the number of solutions of the equation LNX = KX
In this paper, we discuss the number of solutions of the equation LNX = KX
What about the score
When k = 0, the equation becomes LNX = 0, then x = 1, and there is a solution
When K0, Let f (x) = LNX KX, then f '(x) = 1 / x-k = (1-kx) / X,
So f '(x) is greater than 0 on (0,1 / k) and less than 0 on (1 / K, positive infinity),
Therefore, f (x) is monotonically increasing on (0,1 / k) and monotonically decreasing on (1 / K, positive infinity)
So f (x) has the maximum value ln (1 / k) - 1 = - lnk-1 on x = 1 / K
If - lnk-1 > 0, that is K
Drawing
Let the left side be Y1 and the right side be Y2; draw the function image. When K0, the derivative of Y1 is 1 / x, let 1 / x = k, then there is an intersection, in addition, when k > 0, there are two intersections. The number of intersections is equal to the number of solutions of the original equation
Given sin ^ 2 θ (1 + cot θ) + cos ^ 2 θ (1 + Tan θ) = 2, θ∈ (0,2 π), find the value of Tan θ
sin^2 θ(1+cotθ)+cos^2θ(1+tanθ)=2
sin^2 θ(1+cosθ/sinθ)+cos^2θ(1+sinθ/cosθ)=2
sin^2 θ+sinθcosθ+cos^2θ+sinθcosθ=2
2sinθsosθ+1=2
sin2θ=1
θ=π/4
tanθ=1
sin^2 θ(1+cotθ)+cos^2θ(1+tanθ)=2
sin^2 θ(1+cosθ/sinθ)+cos^2θ(1+sinθ/cosθ)=2
sin^2 θ+sinθcosθ+cos^2θ+sinθcosθ=2
2sinθsosθ+1=2
2sinθsosθ=1
2sinθsosθ/(sin^2 θ+cos^2 θ）=1
2tanθ/(tan^2θ+1)=1
It is reduced to (Tan θ - 1) ^ 2 = 0, so tan θ = 1
As shown in the figure, in the plane rectangular coordinate system, the vertex of the rhombic oabc is on the x-axis, and the coordinates of the vertex C are (3,4). If the line L passes through the point (1,0), and the rhombic oabc is divided into two parts with equal area, then the analytical expression of the linear lazy function is (3,4)
What about a good graph, y = 0?