The position relationship between the square of X + the square of Y - 6x = 0 and the square of circle x + the square of Y - 8y = 0. If they intersect, the common chord length can be obtained,

The position relationship between the square of X + the square of Y - 6x = 0 and the square of circle x + the square of Y - 8y = 0. If they intersect, the common chord length can be obtained,

Square of X + square of Y - 6x = 0
(x-3) ^ 2 + y ^ 2 = 9, so the center (3,0), radius 3
The square of X + the square of Y - 8y = 0
X ^ 2 + (y-4) ^ 2 = 16, so the center (0,4), radius 4
So the center distance of the circle is 5, < radius sum, > radius difference, so the two circles intersect. And the common chord length is twice the height of the hypotenuse of the right triangle whose side length is 3,4,5. And the height = 12 / 5, so the common chord length is 24 / 5

The velocity of a particle moving along a straight line is v = T ^ 3 + 3T ^ 2 + 2 (m · s ^ - 1). If x = 4m when t = 2S, the position, velocity and acceleration of the particle when t = 3S are calculated

S = x integral = 1 / 4T ^ 4 + T ^ 3 + 2T; d = s3-s2 = 149 / 4; X3 = X1 + D = 165 / 4
V3=56m/s.
A = vdifferential = 3T ^ 2 + 6T; A3 = 45m / S ^ 2

It is known that the equation of a circle is X & sup2; + Y & sup2; + KX + 2Y + k = 0. If there are two tangent lines of a circle passing through a fixed point P (1, - 1), then K satisfies the condition that

Circle: (x + K / 2) ^ 2 + (y + 1) ^ 2 = (K / 2-1) ^ 2
P has two tangents outside the circle
That is, the distance from P to the center of the circle > radius
（1+2/k）^2>(k/2-1)^2
The solution seems to be k > 0

Given a particle moving in a straight line, its acceleration is a = 4 + 3T M / S2, when it starts to move, x = 5m, v = 0, find the velocity of the particle at t = 8s

What are the standard equation, quasilinear and eccentricity of hyperbola with center at origin and focus on Y axis?

The standard equation y ^ 2 / A ^ 2-x ^ 2 / b ^ 2 = 1, the alignment y = + (-) a ^ 2 / C, the eccentricity e = C / A

Given the motion equation of the particle  r = (3T + 2t) I + (5t-2t) J, we can find the velocity and acceleration when t = 3S

Your title is wrong. It should be 3T ^ 2T and 5T ^ 2
Let me try:
The differential equation of motion r = (3T ^ 2 + 2t) I + (5T ^ 2-2t) J is obtained, v = DR / dt = (6T + 2) I + (10t-2) J
Substituting t = 3S, we get v = 20i + 28J, and V is about 34m / s
For V = DR / dt = (6T + 2) I + (10t-2) J redifferentiation,
A = DV / dt = 6I + 10J, 136 (m ^ 2 / s) under the root sign

For the hyperbola m whose center is at the origin of the coordinate and whose focus is on the x-axis, the eccentricity e is 2, and the distance between the left vertex and the right focus is 6
For the hyperbola m whose center is at the origin of the coordinate and whose focus is on the x-axis, the eccentricity e is 2, and the distance between the left vertex and the right focus is 6
How to find the standard equation of hyperbola m? Given that the circle C passes through the left vertex and right focus of the hyperbola m, and intersects the straight line x + Y-5 = 0, the dazzle length is 2 and the root sign is 11, the equation of circle C is obtained?

E = C / a = 2 and the distance from the left focus to the right vertex = a + C = 6, we can get C = 2, a = 1, so the equation of M is x ^ 2 - (y ^ 2) / 3 = 1

When an object moves in a straight line with uniform acceleration from its rest, the displacement of the object in the second t time is 1.8m
And the velocity at the end of the second t is 2m / s. what is the acceleration of the object?

0.5a*2T*2T-0.5a*T*T=1.8
2aT=2
Simultaneous, t = 1.2, a = 5 / 6
So the acceleration is 5 / 6m / s

It is known that the center of hyperbola is at the origin, the eccentricity is 2, and a focus f (- m, 0) (M is a normal number)
.

F (- m, 0), so C = M
e=c/a=m/a=2
a=m/2,a^2=1/4m^2
b^2=c^2-a^2=m^2-1/4m^2=3/4m^2
Hyperbolic equation: x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1
x^2/(1/4m^2)-y^2/(3/4m^2)=1
4x^2/m^2-4y^2/3m^2=1

When an object starts to move in a straight line with uniform acceleration from its rest, and t is the time interval, the displacement of the object in the second t time is 1.8m, and the velocity at the end of the second t time is 2m / s, then the following conclusion is correct ()
A. The acceleration of the object a = 56m / S2B. The time interval T = 1.0sc. The displacement of the object in the first 3T time is 4.5md. The displacement of the object in the first t time is 0.8m

The instantaneous velocity at the end of 1.5T is equal to the average velocity in the second t time, that is, V1 = 1.8T; from the velocity formula, we can get: 1.8T = a32t; from the velocity formula, we can get: the velocity at the end of 2T: V2 = a2t = 2; from the simultaneous solution, we can get: a = 56m / S2, t = 65 = 1.2s, so a is correct, B is wrong; the displacement in the first 3T time is 12a (3T) 2 = 3.6m