Taking the vertex of ellipse X & # 178 / / A & # 178; + Y & # 178 / / B & # 178; = 1 (a > b > 0) as the focus, the hyperbolic equation with the focus as the vertex is?

Taking the vertex of ellipse X & # 178 / / A & # 178; + Y & # 178 / / B & # 178; = 1 (a > b > 0) as the focus, the hyperbolic equation with the focus as the vertex is?

The focus is on the x-axis
Hyperbola with a ', B', C '
Then a '= C' = a '= C' = C '= C' = C '= C' = C '= C' = C '= C' = C '= C' = C '= C' = C '= C' = C '
c'=a
c²=a²-b²
So a '² = A & #178; - B & #178;
c'²=a²
Then B '² = C' ² - a '² = B & #178;
So it's X & # 178; / (A & # 178; - B & # 178;) - Y & # 178; / B & # 178; = 1
It is known that the vertex of the ellipse coincides with the focus of the hyperbola y24 − X212 = 1, and the sum of their eccentricities is 135. If the focus of the ellipse is on the x-axis, the standard equation of the ellipse is obtained
The elliptic equation is x2a2 + y2b2 = 1, the centrifugal rate is e, the focal length is 2c, the focal length of hyperbola y224 − X212 = 1 is 2C1, the centrifugal rate is E1, the centrifugal rate is E1, the centrifugal rate is E1, the elliptic equation is x2a2 + y2b2 = 1, the centrifugal rate is e, the centrifugal rate is e, the centrifugal rate is e, the focal length is 2c, the focal length of hyperbola y224 − X212 = 1 is 2C1, the centrifugal rate is E1, the centrifugal rate is E1, the centrifugal rate is E1, (2 points) there are: C12 = C12 = 4 = 4 + 12 = 4 + 12 = 4 + 12 = 4 + 12 = 16, C1 = 16, C1 = 4 & amp & nbsp; & nbsp; & & nbsp; & & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & it's not easy; &Nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp;; (4 points) ｜ E1 = C12 = 2 (6 points) ｜ e = 135 − 2 = 35, i.e. CA = 35, ① (8 points) and B = C1 = 4 & nbsp; & nbsp; & nbsp; & nbsp; ② (9 points) A2 = B2 + C2, ③ (10 points) A2 = 25 can be obtained from ①, ② and ③. The obtained elliptic equation is X225 + y216 = 1 (12 points)
The asymptote equation of hyperbola with the focus of ellipse x28 + Y25 = 1 is ()
A. y=±35xB. y=±53xC. y=±155xD. y=±153x
According to the meaning of the title, the focus coordinate of the ellipse x28 + Y25 = 1 is (± 3,0), the vertex coordinate of the hyperbola is (± 3,0), the vertex of the ellipse is the focus of the hyperbola, the focus of the hyperbola is (± 8,0), in the hyperbola, B2 = c2-a2 = 5, and the asymptote equation of the hyperbola is y = ± 153x
Given that the first three terms of arithmetic sequence are a, 4, 3a, the first n terms and Sn = 110, find a and n
2 * 4 = a + 3a4a = 8A = 2A1 = 2A2 = 4a2 = a1 + D4 = 2 + DD = 2An = a1 + (n-1) d = 2 + 2 (n-1) = 2nsn = (a1 + an) * n / 2110 = (2 + 2n) * n / 2n (n + 1) - 110 = 0n ^ 2 + N-110 = 0 (n + 11) (N-10) = 0n = 10 or n = - 11 (rounding, n > 0) so a = 2, n = 10
a+3a=4*2 => a=2
That is, A1 = a = 2, A2 = 4, A3 = 3A = 6
d=a2-a1=2
Sn
= na1+n(n-1)d/2
=2n+n(n-1)2/2
=n^2+n
And Sn = 110
so
n^2+n=110
n^2+n-110=0
(n+11)(n-10)=0
=> n=10
Elementary school Jiangsu Education Press sixth grade mathematics simple operation exercises and answers
12.06＋5.07＋2.94 30.34＋9.76－10.34 ×3÷ ×3 25×7×4 34÷4÷1.7 1.25÷ ×0.8 102×7.3÷5.1 17 + -7 1 － － ,a+b+c=a+ (b + c ),a+b-c=a +(b-c),a-b+c=a –(b-c),a-b-c= a-( b +c);933-15.7-4.3 41.06－19....
Summary of special relativity formula in College Physics
The work of Baidu Library
In the triangle ABC, a, B and C are the opposite sides of the angle a.b.c. the vector M = (2a-c, COSC), n = (CoSb, - b) and M is perpendicular to N. the value of B is obtained
m=(2a-c,cosC),n=(cosB,-b)
∵ m vertical n
∴(2a-c)cosB-bcosC=0
According to the sine theorem:
(2sinA-sinC)cosB-sinBcosC=0
2sinAcosB-(sinCcosB+sinBcosC)=0
2sinAcosB-sin(C+B)=0
2sinAcosB-sinA=0
∵sinA≠0∴cosB=1/2
∵0
Because the vector m is perpendicular to the vector n, there is a
(2a-c)*cosB+cosC*(-b)=0
According to the cosine theorem, CoSb = (a * a + C * C-B * b) / 2Ac; COSC = (a * a + b * B-C * c) / 2Ab
It's too much trouble to simplify
Because m is perpendicular to N, m.n = 0
And the vector M = (2a-c, COSC), n = (CoSb, - b)
(2a-c)cosB+cosC(-b)=0
From the sine theorem a / Sina = B / SINB = C / sinc = 2R
a=2RsinA，b=2RsinB，c=2RsinC
So (4rsina-2rsinc) CoSb + COSC (... Expansion)
Because m is perpendicular to N, m.n = 0
And the vector M = (2a-c, COSC), n = (CoSb, - b)
(2a-c)cosB+cosC(-b)=0
From the sine theorem a / Sina = B / SINB = C / sinc = 2R
a=2RsinA，b=2RsinB，c=2RsinC
So (4rsina-2rsinc) CoSb + COSC (- 2rsinb) = 0
That is 4sinacosb-2sinccosb-2cosc2sinb = 0
2sinAcosB-sin(C+B)=0
2sinAcosB-sinA=0
∵sinA≠0∴cosB=1/2
∵0
Given that the first three terms of arithmetic sequence are a, 4, 3a, the sum of the first n terms is Sn and SK = 90, find the value of a and K
a. 4,3a into arithmetic sequence
a+3a=8
A=2
d=4-a=2
2k+2k(k-1)/2=90
k^2+k-90=0
K = 9 or K = - 10 (rounding)
A=2
K=9
1) 4-A = 3a-4 gives a = 2
Then A1 = 2, d = 4-A = 2
2) The results show that (k) k k k = 1-1 / k k d = 10-1
So k = 9
The operation of number (only the column expression is not calculated) the column synthesis expression
(1) The children's clothing factory plans to make 560 sets of children's clothing in 7 days, but actually only takes 5 days. How many sets of children's clothing are more than the plan? (2) the children's clothing factory plans to make 560 sets of children's clothing in 7 days, but 32 sets more than the original plan. How many days are it completed? (3) the children's clothing factory makes 75 sets of children's clothing every day on average in the first 5 days, and 220 sets of children's clothing every day in the last 3 days?
1.560/5-560 / 7 = 32 sets
560 / (560 / 7 + 32) = 5 days
3.（75*5+220*3）/（3+2）
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Clock slow formula and ruler contraction formula of special relativity
The denominator is 1 - (U / C) ^ 2
So what about u = C photons
Yes. Because of the same frame of reference?
The theory of relativity adapts to the macroscopic high-speed motion, and the macroscopic object cannot reach the speed of light. If it approaches the speed of light, the mass will reach infinity, which is impossible. The static mass of photon is zero, so it is not subject to this limit. The speed of macroscopic object is always less than the speed of light