The moving circle C is inscribed with the fixed circle C1: x ^ 2 + (y-4) ^ 2 = 64, and is circumscribed with the fixed circle C2: x ^ 2 + (y + 4) ^ 2 = 4 The trajectory of C is the trajectory of the center point C, not the circular equation

The moving circle C is inscribed with the fixed circle C1: x ^ 2 + (y-4) ^ 2 = 64, and is circumscribed with the fixed circle C2: x ^ 2 + (y + 4) ^ 2 = 4 The trajectory of C is the trajectory of the center point C, not the circular equation

It is obvious that the locus of point C is the ellipse circle C1: X & sup2; + (y-4) & sup2; = 64, the center (0,4), the radius is 8 circle C2: X & sup2; + (y + 4) & sup2; = 4, the center (0, - 4) radius is 2, the center C is set as (x, y) radius, the sum of the distances from RC to the fixed points (0, - 4) and (0,4) = 8-r + 2 + r = 10
Let C (x, y), the radius of circle C be r (r > 0)
Because the moving circle C is inscribed with the fixed circle C1: x ^ 2 + (y-4) ^ 2 = 64
The distance between C (x, y) and C1 (0,4) is | R-8|
That is: (x-0) ^ 2 + (y-4) ^ 2 = (R-8) ^ 2 -- - (1)
And because: C fixed circle C2: x ^ 2 + (y + 4) ^ 2 = 4 circumscribed
The distance between C (x, y) and C2 (0, - 4) is R + 2
That is: (x-0) ^ 2 + (y + 4) ^ 2... Expansion
Let C (x, y), the radius of circle C be r (r > 0)
Because the moving circle C is inscribed with the fixed circle C1: x ^ 2 + (y-4) ^ 2 = 64
The distance between C (x, y) and C1 (0,4) is | R-8|
That is: (x-0) ^ 2 + (y-4) ^ 2 = (R-8) ^ 2 -- - (1)
And because: C fixed circle C2: x ^ 2 + (y + 4) ^ 2 = 4 circumscribed
The distance between C (x, y) and C2 (0, - 4) is R + 2
That is: (x-0) ^ 2 + (y + 4) ^ 2 = (R + 2) ^ 2 -- - (2)
(2) - (1) get:
(y+4)^2-(y-4)^2=(r+2)^2-(r-8)^2
Then: 16y = 20R
Then: r = 4Y / 5 -- - (3)
(3) Substituting (2) and sorting out:
25x^2+9y^2+120y+300=0
That is, the trajectory equation of C: 25X ^ 2 + 9y ^ 2 + 120Y + 300 = 0
Given the parabola C1: y = x * 2-4x + 3, rotate C1 180 ° around the point P (T, 1) to get C2. If the vertex of C2 is on the parabola C1, find the analytic expression of C2
Given that C1: y = x ^ 2-4 + 3 is transformed into: y = (X-2) ^ 2-1, so the vertex of C1 is (2, - 1). Rotate C1 180 ° around point P (T, 1) to get C2, that is to say, C1 and C2 are symmetric about the center of point P. so the vertex coordinates (a, b) of C2 and the vertex coordinates of C1 satisfy the relationship: (a + 2) / 2 = t (B-1) / 2 = 1, so a = 2t-2, B = 3
Given the parabola C1: y = x ^ 2 + 2x and C2: y = - x ^ 2 + A, if the line L is the tangent of C1 and C2 at the same time, then l is said to be the common tangent of C1 and C2
If there are two common tangents, it is proved that the corresponding two common tangents are equally divided
The line segment between two tangent points on the common tangent line is called the common tangent line segment. In fact, two common tangent lines are equally divided.
Y1 '= 2x + 2, Y2' = - 2x, let the abscissa of Y1 tangent point be p, and substituting it into a parabola, the ordinate is p ^ 2 + 2p. Tangent equation: y-p ^ 2 + 2p = (2P + 2) (X-P), that is, y = (2P + 2) X-2 (P ^ 2) - 2p + P ^ 2-2p = (2P + 2) X-P ^ 2-4p --- (1) let the abscissa of Y2 tangent point be q, substituting it into a parabola, the ordinate is - Q ^ 2 + A. tangent equation: y + Q
Prove (SiNx / x) ^ 2
When 0
Let y = x ^ 2
sinx^2/x^2 = sin(x^2)/x^2 = siny/y
SiNx / X is a decreasing function x > x ^ 2 = y
sinx/x
How to prove SiNx > 2x / π x ∈ (0, π / 2)
Using derivative
f(x)=sinx-2x/π,x∈(0,π/2)
Then f '(x) = cosx - 2 / π, Let f' (x) = 0, then x = arccos (2 / π)
When x ∈ (0, arccos (2 / π)), f '(x) > 0; X ∈ (arccos (2 / π), π / 2), f' (x) 0, the conclusion is valid
Certificate:
Let f (x) = x / SiNx (x ∈ (0, π / 2))
x. SiNx > 0
f'(x)=[sinx+xcosx]/(sinx)^2>0
X / SiNx increases monotonically over (0, π / 2).
When x = π / 2, X / SiNx = π / 2
Therefore, when x ∈ (0, π / 2), X / SiNx is constant, x = arccos (2 / π).. at x = arccos (2 / π), the function f (x) has a maximum, so f (x) min = min = {f (0), f (π / 2)} = 0. When x ∈ (0, π / 2), f (x) > F (0) = 0. That is, SiNx > 2x / π.
Why can SiNx and X compare in size? How?
Because they are all functions. We can draw two images, and the sizes of them are different in different domains
How to compare the size of X and SiNx in the interval (π / 2,0)?
A: is the interval reversed? (0, π / 2) right?
Let f (x) = sinx-x
The derivation is f '(x) = cosx-1
When x, y, Z satisfy x + y + Z = 5, XY + YZ + ZX = 3, find the maximum value of Z
Substituting x = 5-y-z into XY + YZ + ZX = 3,
The quadratic equation y & sup2; + (Z-5) y + Z & sup2; - 5Z + 3 = 0
Because y is a real number, so △≥ 0
That is, (Z-5) & sup2; - 4 (Z & sup2; - 5Z + 3) ≥ 0
The results show that (3z-13) (Z + 1) ≤ 0
The result is - 1 ≤ Z ≤ 13 / 3
So the maximum value of Z is 13 / 3
It is proved that: (1) when x belongs to (0, Pai / 2), sinx-x is cubic / 6
(1) Let f (x) = sinx-x, f '(x) = cosx-1
[sinπ(/2+α)-cos(3π/2-α)]/[tan(2kπ -α)+cot(-kπ+α)]=[sin(4kπ-α)sin(π/2 -α)]/[cos(5π+α)-cos(π/2+α)
This question seems to be wrong. The last part of the question is cos (5 π + α) - cos (π / 2 + α) = 0, and it is the denominator. Please check your input again