Given that the eccentricity of hyperbola is 2 and has the same focus as the ellipse x ^ 2 / 25 + y ^ 2 / 9 = 1, the hypohyperbolic equation is solved

Given that the eccentricity of hyperbola is 2 and has the same focus as the ellipse x ^ 2 / 25 + y ^ 2 / 9 = 1, the hypohyperbolic equation is solved

Ellipse: C = 4
So hyperbola: C = 4, e = C / a = 2 = 4 / A, a = 2, B = 2 √ 3
The hyperbolic equation is: x ^ 2 / 4-y ^ 2 / 12 = 1
Given that the eccentricity of hyperbola x ^ 2 / A ^ 2 &; y ^ 2 / b ^ 2 = 1 is 2, the focus is the same as that of ellipse x ^ 2 / 25 &; y ^ 2 / 9 = 1, then the focus coordinate of hyperbola is; the asymptote equation is
In the ellipse, C & # 178; = A & # 178; - B & # 178; = 25-9 = 16  C = 4, the focus coordinate of hyperbola (- 4,0) (4,0) is the same as that of the ellipse, e = C / a = 2 ∵ C = 4 # a = 2, then B & # 178; = C & # 178; - A & # 178; = 12, B = 2 √ 3
(-√34,0),(√34,0)
Y = ± √ 3x
Given that the eccentricity of hyperbola x2a2 − y2b2 = 1 is 2, the focus is the same as that of ellipse X225 + Y29 = 1, then the equation of hyperbola is______ .
The focal point of ∵ ellipse X225 + Y29 = 1 is F1 (- 4,0), F2 (4,0), ∵ hyperbola x2a2 − y2b2 = 1 is F1 (- 4,0), F2 (4,0), ∵ the eccentricity of hyperbola x2a2 − y2b2 = 1 is 2, ∵ CA = 4A = 2, the solution is a = 2, B = 42 − 22 = 23, ∵ the hyperbola is x24 − y212 = 1
Given sin θ / 2 + cos θ / 2 = 2 √ 3 / 3, find sin θ, Cos2 θ
sinθ/2+cosθ/2=2√3/3
Square on both sides
sin²θ/2+cos²θ/2+2sinθ/2cosθ/2=4/3
1+sinθ=4/3
sinθ=1/3
cos2θ
=1-2sin²θ
=7/9
sinθ/2+cosθ/2=2√3/3
sinθ=2sinθ/2cosθ/2=(sinθ/2+cosθ/2)²-1=4/3-1=1/3
sin²θ=1/9
cos2θ=(1-2sin²θ)=1-2/9=7/9
For example, when the vertex of the square (x + 2, y) = (x + 2, y) in the image passes through the range of the positive axis of the square (x + 2, y) in the analytic formula of the image
&As can be seen from the figure: O (0,0), a (2,0), B (2,2), C (0,2) 1, because the image of the function y = - 3 / 2x2 + BX + C passes through BC two points, substituting the coordinate values of B and C into this function, two equations are obtained: 2 = (- 3 / 2) x2 & # 178; + bx2 + C2 = (- 3 / 2) x0 & # 178; + BX0 + C. solving this equation group, we get: B = 3, C = 2
It is known that there is only one intersection point between the straight line y = kx-1 and the curve X = y & sup2. Find the value of the real number k and the coordinates of this intersection point
1. When y = kx-1, so x = K & sup2; X & sup2; - 2kx + 1K & sup2; X & sup2; - (2k + 1) x + 1 = 0k ≠ 0, there is only one intersection, then the equation has a solution discriminant (2k + 1) & sup2; - 4K & sup2; = 0k = - 1 / 4K = 0y = - 1, parallel to the parabola symmetry axis, there is only one intersection, so k = 0, k = - 1 / 42, k = - 1 / 4, 1 / 1
Y = root x
Substituting into the linear equation, △ = 0, the value of K is obtained
Then the two equations are combined to solve the equations.
Given sin (π / 4 - α) = 5 / 13, α ∈ (0, π / 4), find Cos2 α / cos (π / 4 + α)
When sin (Π / 4 - x) = 5 / 13 √ 2 / 2 (cosx SiNx) = 5 / 13X ∈ (0, Π / 4), cos (Π / 4 - x) = 12 / 13 √ 2 / 2 (cosx + SiNx) = 12 / 13, 1 / 2 (cos ^ 2x sin ^ 2x) = 60 / 169cos2x = 120 / 169cos (Π / 4 + x) = sin (Π / 4 - x) = 5 / 13 (cos2x) / [cos (Π / 4 + x)] = (120 / 169)
The square of both ends is [sin (π / 4-A)] ^ 2 = 25 / 169, that is, [1-cos (π / 2-2a)] / 2 is an acute angle, so cos (2a) = √ [1 - (sin2a) ^ 2] = 120 / 169.
∵α∈(0,π/4)
∴π/4-α∈(0,π/4)
∴cos(π/4-α)=√[1-sin²(π/4-α)]=√[1-﹙5/13﹚²]=12/13
cos2α=sin﹙π/2-2α﹚=sin[2﹙π/4-α﹚]=2sin(π/4-α)cos(π/4-α)
=2×﹙5/13﹚×﹙12/13﹚=120/169
Cos (π / 4 + α) = sin [π... Expansion
∵α∈(0,π/4)
∴π/4-α∈(0,π/4)
∴cos(π/4-α)=√[1-sin²(π/4-α)]=√[1-﹙5/13﹚²]=12/13
cos2α=sin﹙π/2-2α﹚=sin[2﹙π/4-α﹚]=2sin(π/4-α)cos(π/4-α)
=2×﹙5/13﹚×﹙12/13﹚=120/169
cos(π/4+α)=sin[π/2-﹙π/4 ＋α)]=sin(π/4-α)=5/13
∴cos2α/cos(π/4+α)=﹙120/169﹚/﹙5/13﹚
=24 / 13 "put away
Given that the image of a function y = (3-m) x-2m + 10 passes through one, two and four quadrants, the value range of M is
The image of solving the linear function y = (3-m) x-2m + 10 passes through one, two and four quadrants,
Then K < 0 and b > 0
That is, 3-m < 0 and 10-2m > 0
That is, M > 3 and m < 5
That is 3 < m < 5
If the image of a linear function y = (3-m) x-2m + 10 passes through the first, second and fourth quadrants, then K0
Then 3-m0
Then 3
Given that hyperbola x ^ 2 / 9-y ^ 2 / 4 = 1 and straight line y = KX + 1 have an intersection, find the value of K
The asymptote equation of hyperbola x ^ 2 / 9-y ^ 2 / 4 = 1 is y = ± 2x / 3
The straight line y = KX + 1 passes through the fixed point (0,1)
Only the slope of the straight line y = KX + 1 = the slope of the asymptote, the hyperbola x ^ 2 / 9-y ^ 2 / 4 = 1 and the straight line y = KX + 1 have an intersection,
So k = ± 2 / 3
By combining x ^ 2 / 9-y ^ 2 / 4 = 1 with the straight line y = KX + 1, (4-9k & # 178;) x & # 178; - 18kx-45 = 0
When 4-9k & # 178; = 0, that is, k = ± 2 / 3 has an intersection
② When △ k = - 0 ≠
It is known that sin (π / 4 - α) = 5 / 13,0
It is found that [2] / sin a = 2 / 169,
That is [1-cos (π / 2-2a)] / 2 = 25 / 169,
So [1-sin (2a)] / 2 = 25 / 169,
Then sin (2a) = 119 / 169,
Due to 0