Find the elliptic equation which has the common focus with the hyperbola x square - 3Y square = 12 and the sum of the two semiaxes is 8

Find the elliptic equation which has the common focus with the hyperbola x square - 3Y square = 12 and the sum of the two semiaxes is 8

The focus of hyperbola X & # 178 / 12-y & # 178 / 4 = 1 is (- 4,0), (4,0). If C = 4 in the ellipse, then C & # 178; = A & # 178; - B & # 178; = 16 and a + B = 8, then a = 5, B = 3, so the hyperbola is X & # 178 / 25 + Y & # 178 / 9 = 1
The hyperbolic equation with common focus and eccentricity e = 54 is ()
A. x29−y216＝1B. x216−y29＝1C. y29−x216＝1D. y216−x29＝1
∵ the focus of the ellipse x249 + y224 = 1 is (± 5, 0), the hyperbolic equation c = 5, a = 4, B2 = 25-16 = 9, which has a common focus with the ellipse x249 + y224 = 1, and the eccentricity e = 54, is: x216 − Y29 = 1
The eccentricity of ellipse is √ 5 / 3, and the focus of ellipse and hyperbola is the same as that of hyperbola X & # 178 / 4-y & # 178; = 1
Because the ellipse and hyperbola are in common focus, let the elliptic standard equation be x ^ 2 / (4 + k) + y ^ 2 / (k-1) = 1. From e ^ 2 = (C / a) ^ 2 = C ^ 2 / A ^ 2 = (a ^ 2-B ^ 2) / A ^ 2 = 5 / 9, we can get [(4 + k) - (k-1)] / (4 + k) = 5 / 9, and get k = 5. Therefore, the elliptic equation is x ^ 2 / 9 + y ^ 2 / 4 = 1, and its quasilinear equation is x = ± a ^ 2 / C = ± 9
Hyperbola C '= 4 + 1 = 5
Then the ellipse C & # 178; = 5
e²=c²/a²=5/9
a²=9
b²=a²-c²=4
So x & # 178 / 9 + Y & # 178 / 4 = 0
a²=9,c=√5
So the guide line is x = ± 9 √ 5 / 5
Seek primary school sixth grade first volume score four operations of the oral arithmetic problem
[425 -(2.5+1.9) ×0.5]-0.5 1213 -412 -214 -518 -12.5% 0.125×34 +18 ×8.25+12.5% (78 +1316 )÷1316 2.5×37 ×0.4×213 15314 -2....
Thinking about mathematics in the fourth grade volume 2 of people's Education Press
It's a thinking question, not an Olympiad Mathematics question. Be sure to bring the answer. Please!
At least 30 thinking questions, if it's OK,
Hello, the questions are as follows: 1. The boiler room reserves 120 days of heating coal according to the daily consumption of 45 tons. After 40 days of heating, due to the technical transformation, 9 tons of coal are saved every day. How many days can these coal provide heating? 2. There are 16 people in a canteen who can eat rice for 15 days. After 5 days of eating, 6 people leave, and the rest
Question:
In the triangle ABC, the opposite side of! C is B, the vector M = (a + C, B-A) n = (A-C, b) and M is perpendicular to n
M = (a + C, B-A) n = (A-C, b) and M is perpendicular to n
m·n＝0
That is, a ^ 2-C ^ 2 + (B-A) B = 0
a^2+b^2-c^2=ab
cosC=a^2+b^2-c^2/2ab＝1/2
So C = π / 3
You have no problem. What are you asking for
Is it angle c?
From m * n = 0
A * a + b * b-ab = C * C can be obtained
According to the cosine theorem, C = 60 degrees
First of all, we need to know that M is perpendicular to N, and we can deduce an equation: m · n = 0
The vector is known as (a = + B, A-C)
That is, (a + C) (A-C) + B (B-A) = 0
A ^ 2-C ^ 2 + B ^ 2-AB = 0
a^2+b^2-c^2=ab
I don't know what to do when I come here, because you didn't tell me what to ask for... But most of them are for the value of COS, that is, for the angle
First of all, we need to know that M is perpendicular to N, and we can deduce an equation: m · n = 0
Known vector M = (a + C, B-A) n = (A-C, b)
That is, (a + C) (A-C) + B (B-A) = 0
A ^ 2-C ^ 2 + B ^ 2-AB = 0
a^2+b^2-c^2=ab
I don't know what to do here, because you didn't tell me what to ask... But most of them are looking for the value of cos... That is, the size of angle C
cosC=a^2+b^2-c^2/2ab＝1/2
So C = π / 3. So the angle c is 60 degrees
You can ask me any more questions... Put it away
It is known that the tolerance of the arithmetic sequence {an} is not zero, and a 3 = 5, a 1, a 2, a 5 are proportional sequence
Find the sequence {BN} to satisfy B1 + 2B2 + 22b3 + +2n-1bn = an, find the first n terms of sequence {BN} and TN, try to compare TN with (3n-1) / (n + 1)
The sequence {BN} satisfies B1 + 2B2 + 4b3 + +2 ^ (n-1) BN = an
The sequence A1, A2, A5 is the same as the sequence A1, A2, a5
A2 * A2 = A1 * A5 a3-2d = A1 A3 + 2D = A5 a3-d = A2
D=2
b1+2b2+4b3+2^(n-1)bn=an (1)
b1+2b2+4b3+2(n-3)b n-1=a n-1 (2)
(1) - (2) get
2^(n-1)bn=an - a n-1=2
The solution is BN = 2 / 2 ^ (n-1) = 4 / 2 ^ n
b n-1=4/2^(n-1)
bn/b n-1=1/2
Common ratio q = 1 / 2
N = 1 brings in B1 = 2, and the first term is 2
That is, (BN) is equal ratio sequence, TN = 4-2 ^ (3-N)
The second question than the size of your own done, no time!
Who has a set of simple mathematical operations in Volume 2 of grade 4 of primary school?
1.48+9.87+8.52 13.89-4.32-4.68 15.27+4.64-527
1.25×4.6×8 2.5×9.68×0.4 1.25×32×0.25
12×4.5+88×4.5 10.1×9.4 38×0.45+62×0.45
99×0.86 3.6×18-3.6×8
158+262+138 375+219+381+225 5001－247－1021－232
（181+2564）+2719 378+44+114+242+222 276+228+353+219
(375 + 1034) + (966 + 125) (2130 + 783 + 270) + 1017... Expand
158+262+138 375+219+381+225 5001－247－1021－232
（181+2564）+2719 378+44+114+242+222 276+228+353+219
(375+1034)+(966+125) (2130+783+270)+1017 99+999+9999+99999
7755－(2187+755) 2214+638+286 3065－738－1065 899+344
2357－183－317－357 2365－1086－214 497－299 2370+1995
3999+498 1883－398 12×25 75×24
138×25×4 (13×125)×(3×8) (12+24+80)×50 704×25
25×32×125 32×(25+125) 88×125 102×76 58×98
178×101－178 84×36+64×84 75×99+2×75 83×102－83×2
98×199 123×18－123×3+85×123 50×(34×4)×3 25×（24+16）
178×99+178 79×42+79+79×57 7300÷25÷4 8100÷4÷75
16800÷120 30100÷2100 32000÷400 49700÷700
1248÷24 3150÷15 4800÷25 21500÷125
355＋260＋140＋245 102×99 32×125
645－180－245 3600÷4÷25 382×101－382
4×60×50×8 35×8＋35×6－4×35 425+14+186
75+168+25 245+180+20+155 67+25+33+75
60+255+40 548+52+468 13+46+55+54+87
282 + 41 + 159 135 + 39 + 65 + 11 5 + 137 + 45 + 63 + 50
135+38+65+12 30+255+70 280+41+59
8.6-(8.6-3.68)
People's education press fourth grade volume II mathematics first question (page 9) thinking question
.
3-3+3÷3=1
3÷3+3÷3=2
（3+3+3）÷3=3
3÷3+3+3=7
3×3-3÷3＝8
3×3+3-3=9
; female soldier, no; popcorn; gather; just; good opportunity; internationalization; feel good; establish an environment; go home in a few days
At least send out the title?
The third question should be: 3x3-3-3 = 3
3x3 △ 3 △ 3 = 1 3 △ 3 + 3 △ 3 = 2... Expansion
3x3÷3÷3=1 3÷3+3÷3=2 3x3-3-3=3 3÷3+3+3=7 3x3-3÷3=8 3x3x3 △ 3 = 9
In △ ABC, ∠ A and ∠ B are acute angles, and | tanb − 3 | + (2sina − 3) 2 = 0. Try to determine the shape of △ ABC
∵| tanb − 3 | + (2sina − 3) 2 = 0, ∵ tanb = 3, Sina = 32, ∵∵ A and ∵ B are acute angles, ∵ a = 60 °, B = 60 °, C = 180 ° - A - ∵ B = 180 ° - 60 ° - 60 ° = 60 °. ABC is an equilateral triangle