In polar coordinates, the chord length of the intersection of the line psina = √ 2 / 2 and the circle P = 2cosa is

In polar coordinates, the chord length of the intersection of the line psina = √ 2 / 2 and the circle P = 2cosa is

four

The proof of sin α + bcos α = (√ a square + b square) sin (α + θ)

It doesn't need proof, it's just simplification
Asin α + bcos α put forward √ (A & # 178; + B & # 178;)
=√(a²+b²)(√(a²+b²)sinα/a+√(a²+b²)cosα/b)
=√(a²+b²)sin(α+θ) [tanθ=b/a]
for instance
(1).3sinα+4cosα=5(3sinα/5+4cosα/5)=5sin(α+θ) [tanθ=4/3]
(2).1/2sinα+√3cosα/2=sin(α+60°) [tan60°=√3/2÷1/2=√3]

(ACOS α + bsin α) square + (asin α - bcos α) square

The original formula is a & sup2; Cos & sup2; α + 2absin α cos α + B & sup2; sin & sup2; α + A & sup2; sin & sup2; α - 2absin α cos α + B & sup2; Cos & sup2; α
=A²(cos²α+sin²α)+B²(sin²α+cos²α)
=A²+B²

asin α+bcosβ=?

The chord length of the intersection of 2 ρ cos θ = 1 and circle ρ = 2cos θ is ()
A. 3B. 3C. 2D. 1

The line 2 ρ cos θ = 1 is transformed into the ordinary equation: 2x = 1. ∵ ρ = 2cos θ, ∵ ρ 2 = 2 ρ cos θ, which is transformed into the ordinary equation: x2 + y2 = 2x, i.e. (x-1) 2 + y2 = 1. The simultaneous solution is x = 12, y = ± 32, ∵ chord length of the line intersecting with the circle = 3

Finding the chord length of a straight line cut by a circle
What is the chord length of the line X-Y + 4 = 0 cut by the circle x ^ 2 + y ^ 2 + 4x-4y + 6 = 0?

(x+2)²+(y-2)²=2
Center (- 2,2), radius r = √ 2
The center of the circle is on a straight line
So the diameter of chord length is 2 √ 2

Finding the chord length of a circle section line
Finding the known equation of chord length ^ y = 0-2 + X-1

The circular equation becomes: (x-3) ^ 2 + y ^ 2 = 9,
Center C (3,0), radius r = 3,
The distance from the center of the circle to the straight line X-Y + 1 = 0, d = | 3-0 + 1 | / √ 2 = 2 √ 2,
Let the chord be ab,
The chord center distance is the distance from the center of the circle to the straight line. The chord center distance, half chord length and radius form a right triangle,
According to Pythagorean theorem,
R^2-d^2=（AB/2）^2,
∴|AB|=2.

Let X / A + Y / b = 1 pass m (COS a, sin a), then ()
A.a^2+b^2=1
C.1/a^2+1/b^2=1

Choose D
If M (COS a, sin a) is passed, it means that the line is tangent or intersect with the circle, if and only if the distance d from the center of the circle to the line is less than or equal to the radius R
d=1/squr(1/a^2+1/b^2)=1

Given (x / a) cos θ + (Y / b) sin θ = 1, (x / a) sin θ - (Y / b) cos θ = 1, prove (x ^ 2 / A ^ 2) + (y ^ 2 / b ^ 2) = 2
As the title shows

(x/a)cosθ+(y/b)sinθ=1
[(x/a)cosθ+(y/b)sinθ]^2=1
(x/a)sinθ-(y/b)cosθ=1
[(x/a)sinθ-(y/b)cosθ]^2=1
[(x/a)cosθ+(y/b)sinθ]^2+[(x/a)sinθ-(y/b)cosθ]^2=(x^2/a^2)+(y^2/b^2)=2

The curve C: y = sin (7 π / 8-x) cos (x + π / 8) is shifted to the right by a (a 〉 0) units. The obtained curve C 'is symmetric with respect to the straight line x = π / 4
(1) (2) for the minimum value of a, it is proved that when x ∈ (- 8 π / 7, - 9 π / 8), the slope of any two points on the curve C 'is always greater than zero

Y = sin (7 π / 8-x) cos (x + π / 8) = sin (π - (x + π / 8)) cos (x + π / 8) = sin (x + π / 8) cos (x + π / 8) = 1 / 2 * 2Sin (x + π / 8) cos (x + π / 8) = 1 / 2 * sin (2x + π / 4) shifts a (a > 0) units to the right, y = 1 / 2 * sin (2 (x-a) + π / 4) = 1 / 2 * sin (2x-2a + π / 4) straight line x = π / 4 symmetry