Ellipse x ^ 2 / 2m ^ 2 + y ^ 2 / N ^ 2 = 1 and hyperbola x ^ 2 / m ^ 2-y ^ 2 / 2n ^ 2 = 1 have common focus, find the eccentricity of ellipse

Ellipse x ^ 2 / 2m ^ 2 + y ^ 2 / N ^ 2 = 1 and hyperbola x ^ 2 / m ^ 2-y ^ 2 / 2n ^ 2 = 1 have common focus, find the eccentricity of ellipse

We can find that C ^ 2 of ellipse is 2m ^ 2-N ^ 2, and m ^ 2 + 2n ^ 2 of hyperbola. Because there is a common focus, 2m ^ 2-N ^ 2 = m ^ 2 + 2n ^ 2, we can get m ^ 2 = 3N ^ 2. So the square of eccentricity is five sixths. Then we can find the root of eccentricity is 30
If the ellipse x ^ 2 / 10 + y ^ 2 / M = 1 and the hyperbola x ^ 2-y ^ 2 / b = 1 have the same focus, and the ellipse and the hyperbola intersect at (√ 10 / 3, y),
Solving the equation of ellipse and hyperbola
It is known that the eccentricity of ellipse C1: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) is (radical 3) / 2, and the length of line segment of X axis cut by parabola C2: y = x ^ 2-b
The length of the long semiaxis equal to C1
1) The equation of C1, C2
e^2=(a^2-b^2)/a^2=3/4……………… (1)
Because the major semiaxis of C1 = a, C2 is symmetric about the Y axis, and the length of the truncated X axis = major semiaxis a, so C2 = (A / 2,0) is substituted into C2
We get 0 = (A / 2) ^ 2-B （2）
Solution (1) (2) gives a = 2, B = 1
So C1: x ^ 2 / 4 + y ^ 2 = 1
C2：y=x^2-1
In the right triangle ABC, a and B are acute angles, and Tana and tanb are the two roots of the equation 3x ^ 2-tx + 3 = 0,
Sina and SINB are the two roots of the equation x ^ 2-radical 2-k = 0. Find the degree and length of a and B
This is because: CoSb = Sina, SINB = cosasin \35; 178; a + cos \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\178; -
Tana and tanb are the two roots of the equation 3x ^ 2-tx + 3 = 0, Tana * tanb = 3 / 3 = 1
Sina and SINB are the two roots of the equation x ^ 2-radical 2-k = 0, Sina + SINB = radical 2
In the right triangle ABC, a and B are acute angles, SINB = cosa
In addition, Sin & # 178; a + cos & # 178; a = 1sina + cosa = radical 2, we obtain Sina = cosa = radical 2 / 2A = b = 45 degree
It is known that y is a positive proportional function of X. when x increases by 2, the corresponding value of Y decreases by 1, then the proportional coefficient k = ()
Add 20 points to the answer and process within 10 minutes
Let y = KX. Then (Y-1) = K (x + 2), i.e. y = KX + 2K + 1. So 2K + 1 = 0. K = negative 1 / 2
The curvilinear equation is X & # 178; / (K-2) + Y & # 178; / (5-K) = 1. What value range of K is hyperbolic equation? What value of K is elliptic equation
Hyperbolic equation time
k-2>0 -->k>2
5-kk>5
∴k>5
k-2 k0 --> kk>2
5-k>0 -->k
If Tana = 2, then the value of COS (π + a) cos (π / 2 + a) is
A:
tana=2.
cos(π+a)cos(π/2+a)
=(-cosa)*(-sina)
=sinacosa
=sinacosa/[(sina)^2+(cosa)^2]
=1/[sina/cosa+cosa/sina)
=1/(tana+1/tana)
=1/(2+1/2)
=2/5
If Tana = 2, then cos (π + a) cos (π / 2 + a) = - cosa * (- Sina) = sinacosa = sinacosa / (sin ^ 2A + cos ^ 2a)
=tana/(tan^2a+1)=2/5
Given that the proportion coefficient in the positive proportion function y = - x is -? How do you answer?
It's - 1
Given that the equation x ^ 2 / (2 + m) - y ^ 2 / (M + 1) = 1 represents hyperbola, find the value range of M
It's through 2A
Represents a hyperbola
x²/a-y²/b=1
Then a and B have the same sign
So (2 + m) (M + 1) > 0
M-1
Known 0
cos(a+π/4)=cos(a)*cos(π/4) - sin(a)*sin(π/4) =(1/2^0.5)*(cos( a) - sin(a)) = 4/5
cos( a) - sin(a) = (4/5)*2^0.5.(1)
(cos(a+π/4))^2 +(sin(a + π/4))^2 = 1 -> (sin(a+π/4))^2 = 1 - 16/25 = 9/25
0< a sin(a+π/4) = 3/5
sin(a+π/4) = sin(a )*cos(π/4) + cos(a)*sin(π/4) =(1/2^0.5)(cos(a) + sin(a))= 3/5
cos(a) + sin(a) = (3/5)*2^0.5.(2)
(1) + (2) -> 2cos(a) = (7/5)*2^0.5 -> cos(a) = (7/10)*2^0.5
(2) - (1) -> 2sin(a) = (-1/5)*2^0.5 -> sin(a) = (-1/10)*2^0.5
tan(a)= sin(a)/cos(a) = -1/7
Supplement: the problem should be cos (a + π / 4) = - 4 / 5, then the solution is - 7