It is known that the center of the hyperbola is at the origin, the focal points F 1 and F 2 are on the coordinate axis, the eccentricity is 2 ^ 1 / 2 and passes through the point (4, - 10 ^ 1 / 2) It is known that the center of the hyperbola is at the origin, the focus F 1 and F 2 are on the coordinate axis, the eccentricity is 2 ^ 1 / 2 and passes through the point (4, - 10 ^ 1 / 2). Find the hyperbolic equation; if the point m (3, M) is on the hyperbola, find out: the point m is on the circle with the diameter of F 1F 2; find out the area of the triangle f 1m F 2

It is known that the center of the hyperbola is at the origin, the focal points F 1 and F 2 are on the coordinate axis, the eccentricity is 2 ^ 1 / 2 and passes through the point (4, - 10 ^ 1 / 2) It is known that the center of the hyperbola is at the origin, the focus F 1 and F 2 are on the coordinate axis, the eccentricity is 2 ^ 1 / 2 and passes through the point (4, - 10 ^ 1 / 2). Find the hyperbolic equation; if the point m (3, M) is on the hyperbola, find out: the point m is on the circle with the diameter of F 1F 2; find out the area of the triangle f 1m F 2

The centrifugation is 2 ^ 1 / 2
Then a & # 178; = B & # 178;
Let X & # 178; - Y & # 178; = A & # 178;
16-10=a²
a²=6
The equation is X & # 178 / 6-y & # 178 / 6 = 1
c=2√3
M(3,m)
9-m²=6
m²=3
F1M*F2M=(2√3-3,m)*(2√3+3,m)=12-9+m²=0
So f1m ⊥ MF2,
M on a circle with diameter f 1F 2
S=2c*|M|/2=6

When an object starts to move in a straight line with uniform acceleration from a standstill, and t is the time interval, the displacement in the third t is 3M, and the instantaneous velocity at the end of the third t is 3m / s, then ()
A. The acceleration of the object is 1m / S2B. The instantaneous velocity at the end of the first t time is 1m / sc. the time interval is 1.2sd. The displacement of the object in the first t time is 1m

A. The displacement in the third t is x = 12a (3T) 2 − 12a (2t) 2 = 3M. The instantaneous velocity at the end of the third t is v = a · 3T = 3At = 3m / s. The simultaneous solution of two equations leads to t = 1.2s, acceleration a = 56m / S2. So a is wrong, C is correct. B, the velocity at the end of the first t is v = at = 1m / s. so B is correct. D, the displacement in the first t is x = 12at2 = 0.6m. So D is wrong. So BC is selected

Given that the focus of hyperbola is on the x-axis, and a + C = 9, B = 3, then its standard equation is______ ．

Because B = 3, a + C = 9, so c2-a2 = (c + a) (C-A) = 9, so C-A = 1, so C = 5, a = 4, so the standard equation of hyperbola is x216 − Y29 = 1. So the answer is: x216 − Y29 = 1

It is proved that the displacement difference of the object moving in a straight line with uniform speed change is equal in the adjacent equal time interval

s1=Sn-Sn-1=a(nT)^2/2-a{(n-1)T}^2/2=a(2n-1)T^2/2
s2=Sn+1 - Sn=a{(n+1)T}^2/2-a(nT)^2/2=a(2n+1)T^2/2
s3=Sn+2 - Sn+1=a{(n+2)T}^2/2-a{(n+1)T}^2/2=a(2n+3)T^2/2
s2-s1=aT^2
s3-s2=aT^2
So we can prove it

If the focus of the hyperbola is on the X axis and passes through the point m (3,2) n (- 2, - 1), then the hyperbolic standard equation is?

Let the hyperbolic equation be MX ^ 2 + NY ^ 2 = 1
Substituting P and Q into the equation, we can get 9m + 4N = 1,4m + n = 1, and get m = 3 / 7, n = - 5 / 7 by elimination method. So the hyperbolic equation is 3 / 7Y ^ 2-5 / 7X ^ 2 = 1. The standard is y ^ 2 / (√ 21 / 3) ^ 2-x ^ 2 / (√ 35 / 5) ^ 2 = 1
This saves a lot of trouble and is also applicable to ellipses

Why is the difference of displacement equal in the adjacent equal time interval of uniform velocity linear motion
(don't copy from the Internet, I don't know) I want to give a detailed example

Uniform variable speed linear motion means that the speed change is fixed, and the displacement S = speed V * time t in a fixed time. Equal time means that the time t will not change, so the difference of displacement is equal to the difference of speed, and the speed change is fixed, which means that the difference of speed is fixed, so your question has an answer, See if you understand

Find the standard equation of hyperbola with C = √ 6, passing through point (- 5,2) and focus on x-axis

Hyperbola with focus on X-axis
Let X & # 178; / A & # 178; - Y & # 178; / B & # 178; = 1
c=√6
So a & # 178; + B & # 178; = C & # 178; = 6
Passing through point (- 5,2)
So 25 / A and 178; - 4 / B and 178; = 1
Simultaneous solutions of (1) and (2) give a & # 178; = 5 or a & # 178; = 30 (omit)
So B & # 178; = 6-A & # 178; = 1
So the standard equation of hyperbola is X & # 178; / 5-y & # 178; = 1
If you don't understand, please hi me, I wish you a happy study!

In the uniform velocity linear motion, the difference of displacement between any two continuous equal time intervals t is zero_______ Please write down the reasoning process!

The displacements set in time t, 2T, 3T,,,,,, are x1, X2, X3,,,,, X1 = V0, t + at & # 178 / 2x2 = V0 (2t) + a (2t) &# 178 / 2x3 = V0 (3T) + a (3T) &# 178 / 2 respectively, then the displacements S1, S2, S3

Hyperbolic equation with the same focus as hyperbolic standard equation
What is the hyperbolic equation with the same focus as (X & sup2; / A & sup2;) - (Y & sup2; / B & sup2;) = 1?
Is X & sup2; / (A & sup2; + λ) - Y & sup2; / (B & sup2; - λ) = 1?

c²=a²+b²
The same point of intersection is C & sup2; invariant
So x & sup2; / (A & sup2; + λ) - Y & sup2; / (B & sup2; - λ) = 1
Where a & sup2; + λ > 0 and B & sup2; - λ > 0

What is the displacement ratio of uniformly accelerated linear motion with zero initial velocity in continuous equal time interval

The displacement of uniformly accelerating linear motion with zero initial velocity in continuous equal time interval is zero
s1=(1/2)a*t^2
s2=(1/2)a*(2t)^2-(1/2)a*t^2=3*(1/2)a*t^2
s3=(1/2)a*(3t)^2-(1/2)a*(2t)^2=5*(1/2)a*t^2
s4=(1/2)a*(4t)^2-(1/2)a*(3t)^2=7*(1/2)a*t^2
……
Therefore, the ratio of displacement is as follows:
s1:s2:s3:s4…… sn=1:3:5:7:9:…… :2n-1
It's an odd column
Do not know how to test this, odd column is not a physical content. If there are other conditions, also hope to add