Find the hyperbolic standard equation with focus on x-axis, a = 4, B = 3

Find the hyperbolic standard equation with focus on x-axis, a = 4, B = 3

x^2/16-y^2/9=1
x²/16 - y²/9 = 1
If cot θ / (2cos θ + 1) = 1, then Cos2 θ / (1 + sin2 θ) =?
As the title
(COT a) / (2cos a + 1) = 1 cot a = 2cosa + 1 cota-1 = 2cosa cosa Sina = 2sinacosa Sina cosa + 1 = 1-2sinacosa = (Sina COSA) ^ 2 (Sina COSA) ^ 2 - (Sina COSA) - 1 = 0 (Sina COSA) = (1-radical 5) / 2 cosa Sina = (radical 5-1) / 2 (COSA + Sina) ^ 2 = 2 - (...)
Find the hyperbolic equation which is suitable for the following conditions: 1A = 4, B = 3, focus on X-axis 2, focus on (0, - 6), (06) passing through point (2, - 5) 3, focus on
Find a hyperbolic equation suitable for the following conditions:
1A = 4, B = 3, the focus is on the x-axis
2 focus is (0, - 6), (06) passing through point (2, - 5)
3. The focus is on the x-axis, passing through points (negative radical 2, negative radical 3), (one-third radical 15, radical 2)
1、X^/16-Y^2/9=1
2. C = 6, a ^ 2 + B ^ 2 = 36, 25 / A ^ 2-4 / b ^ 2 = 1, the solution is a ^ 2 = 20, B ^ 2 = 16, so the standard equation is y ^ 20-x ^ 2 / 16 = 1
3. Let x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1, then 2 / A ^ 2-3 / b ^ 2 = 1, 5 / 3A ^ 2-2 / b ^ 2 = 1, then a ^ 2 = 1, B ^ 2 = 3
So x ^ 2-y ^ 2 / 3 = 1
[(sin2α/1-cos2α)*(1-cosα/cosα)]*cot(α/2)
Simplification
Two is twice, not square
[(sin2α/1-cos2α)*(1-cosα/cosα)]*cot(α/2)
=[2sinαcosα/(sinα)^2*(1-cosα/cosα)]*cot(α/2)]
=[(1-cosα)/sinα]*(1+cosα)/sinα
=1
I think twice? Not square!
The angle doubling formula is used
sin2α=2sinαcosα;
cos2α=1-(sinα)^2
Half angle formula:
cot(α/2)=(1+cosα)/sinα
Transformation is the answer
A = 4, B = 3, focus on the x-axis, find the hyperbolic standard equation suitable for the condition
solution
a^2+b^2=c^2
therefore
c^2=16+9=25
therefore
The equation of hyperbola is
x^2/16-y^2/9=1
a^2=16 b^2=9
Is cot2 α = sin2 α / Cos2 α?
Wrong. It's just the opposite. It's Cos2 α / sin2 α
.
Unequal
tan2α=sin2α/cos2α
Not equal to cot2 α = Cos2 α / sin2 α
It seems to be
Tangent tan2 α = sin2 α / Cos2 α
The opposite is true
Find the standard equation of hyperbola (1) with focus on X axis, a = 2 √ 5 passing through point a (5, - 2)
(2) After two points a (- 7, - 6 √ 2), B (2 √ 7,3)
(1) So let the hyperbolic equation be (x ^ 2, y = 1 / 2, y = 1 / 2, y = 1 / 2)
Given that sina + cosa = 1 / 5 and 2 / 2 π < a < π, then the value of π of Cos2 is
Given that sina + cosa = 1 / 5 and 2 / 2 π < a < π, then the value of a of Cos2 is
π/2
Hyperbolic equation with common focus and passing through point (3 √ 2,2) with hyperbola x ^ 2 / 16-y ^ 2 / 4 = 1
Let the new equation be x ^ 2 / a-y ^ 2 / (20-a) = 1, substitute the point (3 radical signs 2,2) into the new equation, and the solution is a = 12 or 30. Since it is a hyperbola, a = 12 is obtained by rounding off a = 30, that is, the new equation is x ^ 2 / 12-y ^ 2 / 8 = 1
Given Sina SINB = 4 / 5, Cosa CoSb = 3 / 5, find the value of Cos2 (a-b)
The square of two formulas
sin²a-2sinasinb+sin²b=16/25
cos²a-2cosacoab+cos²b=9/25
And Sin & sup2; X + cos & sup2; X = 1
2-2(cosacoab+sinasinb)==1
cos(a-b)=1/2
cos2(a-b)=2cos²(a-b)-1=-1/2