It is in common focus with the ellipse X & # 178 / 16 + Y & # 178 / 25 = 1 and passes through the point (- 2, √ 10) to find the standard equation of hyperbola

It is in common focus with the ellipse X & # 178 / 16 + Y & # 178 / 25 = 1 and passes through the point (- 2, √ 10) to find the standard equation of hyperbola

So, the ellipse is a ^ 2 ^ 2 + B = 16 ^ 2 / 25,
So C = 3, the focus is F1 (0, - 3), F2 (0,3),
Because P (- 2, √ 10) is on hyperbola,
Therefore, it is defined that in hyperbola 2A = | pf1-pf2 | = | [(- 2-0) ^ 2 + (√ 10 + 3) ^ 2] - √ [(- 2-0) ^ 2 + (√ 10-3) ^ 2]|
=2√5 ,
Therefore, a ^ 2 = 5, B ^ 2 = C ^ 2-A ^ 2 = 4,
Therefore, the hyperbolic equation is y ^ 2 / 5-x ^ 2 / 4 = 1
c²=25-16=9
Let the hyperbolic equation be y & # 178; / (A & # 178;) - X & # 178; / (9-A & # 178;) = 1
Generation (- 2, √ 10): 10 / A & # 178; - 4 / (9-A & # 178;) = 1
The solution is: A & # 178; = 5 or a & # 178; = 18
The standard equation is: Y & # 178 / 5-x & # 178 / 4 = 1, let a > 1, then the hyperbola X & # 178 / A & # 178; - y... is expanded
c²=25-16=9
Let the hyperbolic equation be y & # 178; / (A & # 178;) - X & # 178; / (9-A & # 178;) = 1
Generation (- 2, √ 10): 10 / A & # 178; - 4 / (9-A & # 178;) = 1
The solution is: A & # 178; = 5 or a & # 178; = 18
The standard equation is: Y & # 178; / 5-x & # 178; / 4 = 1 question: let a > 1, then the range of eccentricity e of hyperbola X & # 178; / A & # 178; - Y & # 178; / (a + 1) &# 178; = 1 is
The elliptic standard equation with a major axis of 8 has the same focus as the hyperbola 3x & # 178; - Y & # 178; = 3
The hyperbolic equation is reduced to x ^ 2-y ^ 2 / 3 = 1,
Since the ellipse and hyperbola have the same focus, let the elliptic equation be x ^ 2 / (1 + k) + y ^ 2 / (K-3) = 1 (k > 3),
So a ^ 2 = 1 + K,
Because the length of major axis is 2A = 8, so a = 4, so 1 + k = a ^ 2 = 16,
The solution is k = 15,
Therefore, the elliptic equation is x ^ 2 / 16 + y ^ 2 / 12 = 1
It is known that the tolerance of the arithmetic sequence {an} is not 0, A1 = 1 and A1, A3 and A9 are equal proportion sequence. (1) find the general term formula an. (2) let BN = 2 & nbsp; an, find the first n term and Sn of the sequence {BN}
(1) The tolerance D ≠ 0, A1 = 1 and A1, A3, A9 are equal ratio sequence, we get: (1 + 2D) 2 = 1 + 8D, the solution is d = 1 or D = 0 (rounding off), so the general term of {an} is an = n. (2) ∵ BN = 2 & nbsp; an = 2n, the first n terms of {BN} and: SN = 2 + 22 + +2n=2(1−2n)1−2=2n+1-2．
Mathematics problems of grade six
A fraction, its denominator plus 3 can be divided into 3 / 7, its denominator minus 2 can be divided into 2 / 3, what is the fraction?
Requirements: detailed, accurate,
Because adding 3 or subtracting 2 to the denominator, the numerator does not change,
Because 3 / 7 = 6 / 14, 2 / 3 = 6 / 9
So 6 is the least common multiple of the numerator, 14 is the least common multiple of the denominator plus 3, and 9 is the least common multiple of the denominator minus 2, so the denominator is 11 and the numerator is 6
What is Einstein's formula of special relativity?
The formula of relative time is: Δ t = Δ t0 / √ (1-V ^ 2 / C ^ 2); the formula of relative length is: l = l0 √ (1-V ^ 2 / C ^ 2). These two formulas are derived from Lorentz transformation. For the original formula and derivation of Lorentz transformation, please refer to "narrow
In △ ABC, the ABC angle corresponds to the ABC edge, and the vector M = (COS C / 2, sin C / 2), the vector n = (COS C / 2, - Sin C / 2), and the vector m multiplies the vector n
It is known that the vector M = (COS C / 2, sin C / 2), the vector n = (COS C / 2, - Sin C / 2), and the vector m times the vector n is equal to 1 / 2
(1) Find angle c
(2) If C = 7 / 2 area of triangle ABC s = 3 root 3 / 2 find a + B
(1) M. n = 1 / 2 (COS C / 2, sin C / 2); (COS C / 2, - Sin C / 2) = 1 / 2 (COS C / 2) ^ 2 - (sinc / 2) ^ 2 = 1 / 2cos2c = 1 / 22c = 60 ° or 300 ° C = 30 ° or 150 ° (2) C = 7 / 2, triangle ABC area s = 3 √ 3 / 2to find:a+bS = (1/2)absinCab =3√3/2 By...
Let the sum of the first n terms of the arithmetic sequence {an} be Sn, and A1 = 2, A3 = 6, (1) find the general formula of the sequence {an}; (2) if SK = 110, find the value of K
(3) The sum of the first n terms of the sequence {1 / Sn} is TN. find the value of t2013
1D = (a3-a1) / (3-1) = 2An = 2 + 2 (n-1) = 2n2sk = K (2 + 2K) / 2 = K (1 + k) = 110k = 10 or K = - 11 (rounding) 31 / Sn = 1 / N (1 + n) = (1 / n) - [1 / (1 + n)] t2013 = 1-1 / 2 + 1 / 2-1 / 3 + 1/2013-1/2014=1-1/2014=2013/2014
Mathematics sixth grade volume two knowledge and ability training number operation (3), how to write the first question
Students send up the topic
The kinetic energy formula of special relativity is 1 - (V ^ 2 / C ^ 2) - 1) MC ^ 2 under 1 / radical. When V is very small, the kinetic energy formula is 0.5mv ^ 2
When V is small enough, how can we wait from the first formula to the second formula? If yes, we will add >
Ek=m0/√(1-v²/c²)c²-m0c²
Lim(Ek,c→+∞)=m0v²/2
or
c=kv
Ek=m0/√(1-v²/(kv)²)(kv)²-m0(kv)²
Lim(Ek,k→+∞)=m0v²/2
Taylor series
F(A)=SIGM[I=0 TO OO]((F(X0)(I))/I!)+F(X0)
Ignore the high order small quantity
You're talking about the energy formula, not the kinetic energy formula. EK = E-E '= MC2 / √ (1-v2 / C2) - MC2 is the kinetic energy formula. As for how to turn 1 / 2mV ^ 2 into 1 / 2mV ^ 2, it's difficult for mobile phones to explain. I'm sorry to ask: small neglect? Is it the limit?
In △ ABC, the opposite sides of angle A.B.. C are a.b.c.cos (a + C / 2) = √ 3 / 3
In △ ABC, the opposite sides of angle A.B.. C are a.b.c.cos (a + C / 2) = √ 3 / 3. Find the value of CoSb 2. Find the vector Ba * vector BC = 2. B = 2 √ 2. Find the value of a and C
cos(A+C)/2=sin(90-(A+C)/2)=sinB/2=√3/3cosB=1-2sin(B/2)^2=1-2/3=1/3sinB=2√2/3BA*BC=cacosB=2 ca=6b^2=a^2+c^2-2accosB=a^2+c^2-4a^2+c^2=(2√2)^2+4=12ac=6(a+c)^2=a^2+c^2+2ac=12+12=24a+c=2√6a^2-2√6a+6=...