Find the elliptic equation of two foci passing through point a (3,4) and hyperbola X & # 178 / 6-y & # 178 / 3 = 1 Wrong. It's not an ellipse, it's a circle

Find the elliptic equation of two foci passing through point a (3,4) and hyperbola X & # 178 / 6-y & # 178 / 3 = 1 Wrong. It's not an ellipse, it's a circle

a²=6,b²=3
Then C & # 178; = 9
C=3
So the focus is on F1 and F2 (± 3,0)
The center of the circle is on the vertical bisector of F1F2
That is, x = 0
Let C (0, m) be the center of the circle
Then CF1 & # 178; = Ca & # 178;
So 3-178; + m-178; = 3-178; + (M-4) 178;
9+m²=9+m²-8m+16
M=2
So R & # 178; = AC & # 178; = 13
x²+(y-2)²=13
What do you want to ask? Follow up: for the equation of circle, someone else gave you a no answer gold medal
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
Find the standard equation of hyperbola with the focus of ellipse x ^ 2 / 4 + y ^ 2 / 12 = 1 as the vertex and the focus of ellipse on y as the vertex
ellipse
a²=12,b²=4
So C & # 178; = 8
So the hyperbola is a '² = C & #178; = 8, C' ² = A & #178; = 12
So B '² = 4
So y & # 178 / 8-x & # 178 / 4 = 1
In triangle ABC, we know that cosa + CoSb + COSC = 3 / 2, and prove that triangle ABC is equilateral triangle by vector
Vector method Oh~
In the arithmetic sequence {an} and proportional sequence {BN}, A1 = B1 = 1, B4 = 8, the first 10 terms of {an} and S10 = 55. (I) finding an and BN; (II) finding the sum of the first n terms of CN with CN = an + BN
(I) let the tolerance of arithmetic sequence be D, the common ratio of arithmetic sequence be Q. ∵ A1 = B1 = 1, B4 = 8, the first ten terms of {an} and S10 = 55. ∵ S10 = 10 + 10 × 92d = 55; B4 = Q3 = 8; the solution is D = 1, q = 2. So: an = n, BN = 2N-1. (II) ∵ an = n, BN = 2N-1, ∵ CN = an + BN = n + 2N-1, ∵ CN = an + BN = n + 2N-1, TN = (1 + 2 + 3 + +n）+（1+2+4+… +2n-1）=n(n+1)2+1−2n1−2=n(n+1)2+2n−1．
1、 7.42 × 20.15
2、 (3.3 × 64.8 × 50) / (111 × 32.4 × 5)
7.42×20.15
=(7.5-0.08)*(20+0.15)
=149.513
or
7.42×20.15
=7×1.06×20.15
=7×(1+0.06)×(20+0.15)
=7×21.359
=149.513
（3.3×64.8×50）÷（111×32.4×5）
=(3.3÷111)×(64.8÷32.4)×(50÷5)
=0.66
1、 Answer 149.513
2、 Answer 0.594594595
o `~Oo~~
7.42×20.15
=(7.5-0.08)*(20+0.15)
=149.513
（3.3×64.8×50）÷（111×32.4×5）
=(3.3÷111)×(64.8÷32.4)×(50÷5)
=0.66
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Find the formula related to special relativity and the meaning of letters in the formula
This is mainly about Lorentz transformation. It's not easy to write here. Let's look for a book. It's suggested to take a look at Zhao Kaihua's new concept physics, which is relatively simple and involves a lot of physical meanings. It's easy to learn
In △ ABC, the opposite sides of angles a, B and C are a, B and C respectively, if the vector M = (CoSb, COSC) and the vector n = (2a-b, c) are collinear（
In △ ABC, the opposite sides of angles a, B and C are a, B and C respectively. If the vector M = (CoSb, COSC) and the vector n = (2a-b, c) are collinear (1) find the size of ∠ C (2) if C = 2 times the root sign 3, find the maximum value of a + B
(1) ∵ the common line of vector m and vector n ∵ CoSb / 2a-b = COSC / C, namely CoSb. C - (2a-b). COSC = 0
In △ ABC, from the sine theorem: A / Sina = B / SINB = C / sinc = k ≠ 0
∧ a = ksina, B = ksinb, C = ksinc
K[ cosB.sinC -2sinAcosC+sinBcosC]=k[sin（B+C)-2sinAcosC]=ksinA(1-2cosC)=0
∵ ABC ∈ (0, π), ∵ Sina ≠ 0, 1-2cosc = 0, that is, COSC = 1 / 2, the solution is C = π / 3
(2) From (1), k = C / sinc = 4
a+b=ksinA+ksinB=4[sin(2π/3 - B)+sinB]=4√3sin(π/3+B)
∵ B ∈ (0,2 π / 3) ∵ when B = π / 3, there is a + B (max) = 4 √ 3
It is known that {an} is an arithmetic sequence, {BN} is an equal ratio sequence, and A1 = B1 = 2, B4 = 54, a1 + A2 + a3 = B2 + B3. (1) find the general term formula of {an} and {BN}; (2) find the first n term and Sn of {CN} if {CN} satisfies CN = anbn
(1) Let the tolerance of {an} be D, and the common ratio of {BN} be Q. from B4 = b1q3 = 54, Q3 = 542 = 27, thus q = 3, so BN = B1 & nbsp; · & nbsp; QN − 1 = 2 & nbsp; · & nbsp; 3N − 1 (3 points) and a1 + A2 + a3 = 3a2 = B2 + B3 = 6 + 18 = 24, A2 = 8, thus d = a2-a1 = 6, so an = a1 + (n-1) · 6 =
The simple calculation of 0.24 × 1.9 + 2.4 × 0.81
0.24×1.9+2.4×0.81
=2.4*0.19+2.4*0.81
=2.4*（0.19+0.81）
=2.4*1
=2.4
0.24×1.9+2.4×0.81
=0.24×1.9+0.24×8.1
=0.24×（1.9+8.1）
=0.24×10
=2.4
Replace 0.24 with 2.4
=0.24×1.9+0.24×8.1
=0.24×（1.9+8.1）
=0.24×10
=2.4
Amount 2.4x0.19 + 2.4x0.81 = 2.4x (0.19 + 0.81) = 2.4x1 = 2.4
0.24×1.9+2.4×0.81
=0.24×1.9+0.24×8.1
=0.24×（1.9+8.1)
=0.24×10
=2.4