Hyperbola x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1 (a > 0, b > 0) f is the right focus, P is a point on the right branch of the hyperbola, P is above the X axis, and M is a point on the left quasilinear O is the origin of coordinates. OmpF is a parallelogram, | PF | = λ| of | (1) Find the relationship between eccentricity e and λ of hyperbola (2) If | ab | = 12, the hyperbolic equation at this time can be obtained (troublesome steps)

Hyperbola x ^ 2 / A ^ 2-y ^ 2 / b ^ 2 = 1 (a > 0, b > 0) f is the right focus, P is a point on the right branch of the hyperbola, P is above the X axis, and M is a point on the left quasilinear O is the origin of coordinates. OmpF is a parallelogram, | PF | = λ| of | (1) Find the relationship between eccentricity e and λ of hyperbola (2) If | ab | = 12, the hyperbolic equation at this time can be obtained (troublesome steps)

(1) Let the right focus f coordinate of hyperbola be (C, 0), C > 0, and the left focus be point E, then:
|OF|=c,|PF|=λ|OF|=λc
According to the (first) definition of hyperbola:
|PE|-|PF|=2a
The results show that PE | = | PF | + 2A = λ C + 2A
Because PM / / of is PM / / X axis, and the left guide line is perpendicular to the X axis
So PM is perpendicular to the left alignment
That is, the distance from point P to the left quasilinear is the length of the line segment | PM |
In the parallelogram OmpF, there are:
|PM|=|OF|=c
From the second definition of hyperbola, we can get that
|PE|/|PM|=c/a
Then E = C / A
=|PE|/|PM|
=(λc+2a)/c
＝λ+2a/c
=λ+2/e
So λ = E-2 / E
(2) Since point m is on the left quasilinear x = - A & # / C,
So we can set the coordinate of point m as (- A & # / C, y), point P (x, y), where x > 0, Y > 0
Then | Mo | = √ [(- A & # 178 / / C) &# 178; + Y & # 178;]
If λ = 1, then | PF | = | Mo | = | of | = C
So √ [(- A & # 178 / / C) &# 178; + Y & # 178;] = C
That is Y & # 178; = C & # 178; - (A & # 178 / C) &# 178; (1)
It is also known from the first sub problem that λ = E-2 / E
Because λ = 1, so
e-2/e＝1
That is E & # 178; - E-2 = 0
(e-2)(e+1)=0
Because E > 1, the solution is e = 2
Then C / a = 2 and C = 2A
If B & # 178; = C & # 178; - A & # 178;, then B = √ 3a
Hyperbolic equation can be reduced to: 3x & # 178; - Y & # 178; = 3A & # 178; (2)
By substituting C = 2A into (1), we get that:
y²＝4a²-(a²/2a)²=4a²-a²/4=15a²/4
The solution is y = √ 15A / 2
Then the point P coordinate can be expressed as (x, √ 15A / 2)
If point P is on the hyperbola, then the coordinate of point P is substituted into the hyperbolic equation (2)
3x²-15a²/4=3a²
3x²=27a²/4
The solution is x = 3A / 2
Then the point P coordinates are (3a / 2, √ 15A / 2)
So the slope of the straight line OP K (OP) = (√ 15A / 2) / (3a / 2) = √ 15 / 3
That is, the slope of the line passing through the focus f (2a, 0) and parallel to OP is also equal to √ 15 / 3
Then, from the point oblique equation of the straight line, it is obtained that:
y-0=√15/3 *(x-2a)
That is y = √ 15 / 3 * (x-2a)
√15x-3y-2√15 a=0
The linear equation and hyperbolic equation 3x & # 178; - Y & # 178; = 3A & # 178; are combined to find the A.B coordinates of the intersection point
y=√15/3 *(x-2a)
3x²-y²=3a²
The elimination of Y leads to: 3x & # 178; - 5 / 3 * (x-a) &# 178; = 3A & # 178;
That is 9x & # 178; - 5x & # 178; + 10ax - 5A & # 178; = 9A & # 178;
4x²+10ax－14a²＝0
2x²+5ax－7a²＝0
(2x+7a)(x-a)=0
The solution is X1 = - 7a / 2, X2 = a
Because | ab | = √ (1 + K & # 178;) * | x1-x2 | = 12
So √ (1 + 5 / 3) * | - 7a / 2-A | = 12
(2√6)/3 *9a/2=12
The solution is a = (2 √ 6) / 3
Then B = √ 3, a = 2 √ 2
So the hyperbolic equation is:
x²/(8/3) - y²/8=1

The circle O AB is the diameter of the circle AC is the chord D. on the chord AC, OD = 5 ∠ AOD = 2 ∠ a = 60 ° find the length of CD
The value of ∠ AOD = 2 ∠ a = 60 ° should be ∠ ADO = 2 ∠ a = 60 °

From the question ∠ DOA = 90 °, Δ doa is a right triangle
So Ao = 5 × Tan 60 ° = 5 √ 3, ad = 5 / cos 60 ° = 10
So the diameter is ab = 2oa = 10 √ 3
Connect CB, because AB is the diameter, so Δ ACB is a right triangle
AC=AB×cos∠CAB=10√3×cos30°=15
So DC = ac-ad = = 15-10 = 5

Given that AB is the diameter of ⊙ o, point D is on the chord AC, and OD = 5cm, ∠ ADO = 60 ·, do ⊥ AB is in O, find the length of CD
Urgent····

If ∠ ADO = 60 ° do ⊥ AB is in O, then ∠ Dao = 30 ° ad = 2 (OD) = 2 × 5 = 10 (CM) [the right side opposite 30 degrees = half of the hypotenuse];
AO²=AD²-OD²=10²-5²=75,
Connect BC, then ∠ ACB = 90 °, Dao = 30 °, cab = 30 °, BC = AB / 2 = 2ao / 2 = Ao,
AC²=AB²-BC²=(2AO)²-AO²=4AO²-AO²=3AO²=3×75=225,
AC=15(cm),
CD=AC-AD=15-10=5(cm)
Or ∠ ADO = 60 ° do ⊥ AB is in O, then ∠ Dao = 30 ° ad = 2 (OD) = 2 × 5 = 10 (CM) [the right side opposite 30 degrees = half of the hypotenuse];
AO²=AD²-OD²=10²-5²=75,
If BC is connected, then ∠ ACB = 90 °, Dao = 30 °= ∠ cab = 30 °,
Rt△AOD∽Rt△ACB (AAA)
AO:AC=AD:AB
AC=AO*AB/AD=AO*2AO/AD=2AO²/AD=2*75/10=15(cm)
CD=AC-AD=15-10=5(cm)

As shown in the figure, it is known that AB is the diameter of circle O, point D is on the chord AC, and OD = 5cm ∠ ADO = 60 ° do is perpendicular to AB to find the length of CD

If ∠ ADO = 60 °, do ⊥ AB is in O, then ∠ Dao = 30 °, ad = 2 (OD) = 2 × 5 = 10 (CM) [the right side opposite 30 degrees = half of the hypotenuse]; AO = ad-od = 10-5 = 75, connecting BC, then ∠ ACB = 90 °, Dao = 30 ° = ∠ cab = 30 °, BC = AB / 2 = 2ao / 2 = Ao, AC = ab-bc = (2ao) - Ao = 4ao-ao = 3AO = 3 × 75 = 225, a

O is a point on the straight line AB, and the angle AOC = 53 degrees od bisects the angle BOC to find the degree of BOD

∵ o is a point on the line ab
∴∠BOC＝180-∠AOC＝180-53＝127
∵ od bisection ∠ BOC
∴∠BOD＝∠BOC/2＝127/2＝63.5°
The math group answered your question,

O is a point on the straight line AB, OC, od are two rays from O, OE bisects angle AOC, angle BOC: angle AOE: angle AOD = 2:5:8, and finds the degree of angle BOD

Solution to AOC of known OE bisector angle
Then angle AOE = angle Coe
Because ∠ AOE + ∠ eco + cob = 180 degree
And because ∠ cob; ∠ AOE = 2; 5
So 2A + 5A + 5A = 180
12a=180
a=1 5
Then ∠ AOD = 15 × 8 = 120 degree
Therefore, DOB = 180 minus 120
=60°

When the known angle BOC is outside the angle AOB, the OE bisector angle AOB, the of bisector angle BOC, the OD bisector angle AOC, the angle AOE = 30 degrees, and the angle BOD = 20 degrees are calculated
Degree of angle COF

We know that OE bisector angle AOB, angle AOE = 30 degrees, so angle EOB = angle AOE = 30 degrees
Angle BOD = 20 degree angle AOD = angle AOE + angle EOB + angle BOD = 80 degree
Because od bisects angle AOC, so angle doc = angle AOD = 80 degrees
Angle BOC = angle BOD + angle doc = 20 + 80 = 100 degrees
Because of bisects the angle BOC, the angle COF = 100 / 2 = 50 degrees

It is known that the angle AOB is 150 degrees, OC is a ray in the angle AOB, the ray od bisects the angle AOC, and the ray OE bisects the angle BOD
(1) If AOD = EOC, calculate the degree of AOD
(2) If angle AOD = a (a is not equal to 50 degrees), find | angle AOD - angle BOE | / angle Coe
15 minutes

1、
∵ od bisection ∠ AOC
∴∠AOC＝2∠AOD
∵ OE bisection ∠ BOC
∴∠BOC＝2∠EOC
∴∠AOD＝∠EOC
∴∠AOC＝∠BOC
∴∠AOB＝∠AOC+∠BOC＝2∠AOC
∴2∠AOC＝150
∴∠AOC＝75
∴∠AOD＝∠AOC/2＝37.5
2、
∵ od bisection ∠ AOC
∴∠AOC＝2∠AOD＝2a
∴∠BOC＝∠AOB-∠AOC＝150-2a
∵ OE bisection ∠ BOC
∴∠BOE＝∠COE＝∠BOC/2＝（150-2a）/2＝75-a
∴｜∠AOD-∠BOE｜/∠COE＝｜a-75+a｜/（75-a）＝75/（75-a）
The math group answered your question,

OC / OD is the two rays in ∠ AOB, OE bisects ∠ AOC, of bisects ∠ BOD
(1) If ∠ AOB = 140 °, COD = 30 °, calculate the degree of ∠ EOF
(2) The value of ∠ EOF = α, ∠ cod = β is calculated as ∠ AOB

(1)∠EOF=∠EOC+∠COD+∠DOF∠AOC+∠BOD=∠AOB-∠COD=140-30=110∠EOC+∠DOF=1/2(∠AOC+∠BOD)=55∠EOF=30+55=85(2)∠EOF=∠EOC+∠COD+∠DOF=∠EOC+∠DOF+β=α∠EOC+∠DOF=α-β∠EOC+∠DOF=1/2(∠AOC+∠BOD)∠AOC+...

As shown in the figure, ray OC and OD divide AOB into three equal parts, OE equal parts ∠ AOC, of equal parts ∠ BOD
1. Calculate the degree of ∠ cod
2. Write all the right angles in the diagram
3. Write all the remaining angles and complements of ∠ cod

(1) AOB = 180 ° OC and OD divide AOB into three equal parts, ∠ cod = ∠ AOC = ∠ BOD = 1 / 3 ∠ AOB = 60 °, and (2) OE and of divide AOC and BOD equally, and ∠ COE = 1 / 2 ∠ AOC = 30 °, DOF = 1 / 2 ∠ BOD = 30 °, and ∠ DOE = ∠ COF = 90 °, i.e., ∠ DOE and ∠ COF are right angles