The coordinates of the focus are (- 6,0), (6,0) and the standard equation of hyperbola passing through point a (- 5,2)

The coordinates of the focus are (- 6,0), (6,0) and the standard equation of hyperbola passing through point a (- 5,2)

The focus of the hyperbola is on the y-axis. Let the equation of the hyperbola be y & # 178 / A & # 178; - X & # 178 / B & # 178; = 1
If two focal coordinates (0, c), (0, - C) are (0,6), (0, - 6), then C = 6
From a & # 178; + B & # 178; = C & # 178; there is a & # 178; + B & # 178; = 36 to get B & # 178; = 36-a & # 178;
The substitution points (2, - 5) have 25 / A and 178; - 4 / B and 178; = 1
25/a²-4/(36-a²)=1
25(36-a²)-4a²=a²(36-a²)
After sorting out and decomposing the factors, we get (A & # 178; - 20) (A & # 178; - 45) = 0
If a & # 178; = 20, (B & # 178; = 36-45 < 0, then a & # 178; = 45) is omitted
b²=c²-a²=36-20=16
The equation of the hyperbola is Y & # 178 / 20-x & # 178 / 16 = 1

The ratio of adjacent displacements at equal time intervals for uniform acceleration motion with zero initial velocity

This is based on △ s = at ^ 2!
Ratio of displacements in equal time
That is to say, the displacement ratio is only 1:3:5:9: (2n-1) in 1s, 2S and 3S
·-·-·-·-·-·-·-·-·-·-·-·--·-·-·-·-·--·
The displacement ratio of 1s, 2S and 3S is 1:4:9:...: n ^ 2

Focus on (0, - 6), (0,6), through the point (2, - 5) to find the standard equation of hyperbola? To reason!
The focus is (0, - 6), (0,6), passing through points (2, - 5)
How to find the standard equation of hyperbola?

And C = 6 sets up Y-axis and C = 6, and C = 6, y-y-\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\(a) &

When a particle moves in a variable speed linear motion, its unique function is s (T) = t + 1 / T. the velocity function v (T) and acceleration function a (T) are obtained

The first derivative of velocity function v (T) s (T) to t s (T) = 1-1 / T ^ 2
Second derivative of acceleration function a (T) s (T) to t s (T) = 2 / T ^ 3

Find the hyperbolic standard equation, the two focus coordinates are (0, - 6) (0,6) and pass through the point (2, - 5)

Let x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1 focus (0, - 6) (0,6) so a

When a particle moves along a circle with radius r, the initial velocity V0 and the normal acceleration are always equal to the tangential acceleration, and the velocity V of the particle passing through time t is calculated?

The answer is shown in the picture

Find the hyperbolic equation with the focus of the ellipse 16 / x square + 25 / y square = 1 as the vertex and the vertex as the focus

C = 5, a = 3, B = 4 of hyperbola with vertex (0, - 5), (0,5) of ellipse x ^ 2 / 16 + y ^ 2 / 25 = 1 as focus and vertex (0, - 3), (0,3)
So the standard equation of hyperbola is y ^ 2 / 9-x ^ 2 / 16 = 1

When a particle moves in a curve, the motion function is r = Ti + T ^ 2J, and the tangential acceleration at and the normal acceleration an at time t are calculated

&The equation of motion is r = Ti + T ^ 2J velocity vector V = I + 2tj acceleration vector a = 2J tangential velocity

The focus and vertex of an ellipse are the vertex and focus of hyperbola (y squared / 16) - (x squared / 9) = 1, respectively

The vertex of the hyperbola is (0,4), (0, - 4), and the focus is (0,5), (0, - 5)
So, in the ellipse, a = 5, C = 4
Then: B & # 178; = 9
Therefore, the elliptic equation is: Y & # 178 / 25 + X & # 178 / 9 = 1

When a car moves in a straight line, the speed is V0 when braking, and the relationship between acceleration and speed after braking is a = - kV, K is known. The relationship between car speed and time after braking can be obtained

This is a calculus problem, because the mathematical integral symbol can not be written, here is abbreviated excuse me
A = DV / DT, so - kV = DV / DT, 1 / V DV = - K DT, integral on both sides, left side from v0 to V, right side from 0 to t, [ln v] V0 -- v = - KT, so ln v-ln V0 = - KT, LNv = lnv0 KT