Polar coordinate equation of conic Polar coordinate equation of ellipse ｛ 61554; y = EP / (1-ecosa ｛ 61553;) (00 is focal parameter) The polar equation of hyperbola (E > 1, P > 0 as focal parameter) In this equation, y is r, what if it is positive or negative? Is it a part of conic? What about the equation in polar coordinate system established in other ways?

(1) The eccentricity is 0.5, and the distance from focus to guide line is 6
(2) The major axis is 10 and the minor axis is 8
The polar coordinate equation of an ellipse ﹤ 61554; ρ = EP / (1-ecos θ & ﹤ 61553;) (0 ﹤ e ﹤ 1, P is the distance from the focus to the directrix)
therefore
(1) The eccentricity is 0.5, and the distance from focus to guide line is 6
ρ=0.5*6/(1-0.5cosθ）=3/(1-0.5cosθ）=6/(2-cosθ）
(2) The major axis is 10 and the minor axis is 8
Then a = 10 / 2 = 5, B = 8 / 2 = 4
So C = 3
So e = 3 / 5 = 0.6, P = A & sup2 / / C-C = 5 & sup2 / / 3-3 = 16 / 3
So ρ = (0.6 * 16 / 3) / (1-0.6cos θ) = 3.2 / (1-0.6cos θ) = 16 / (5-3cos θ)
Given that the polar coordinate equation of curve C is: ρ 2-2 ρ cos θ - 4 ρ sin θ + 4 = 0, the Cartesian coordinates of any point P on curve C is (x, y), then the value range of 3x + 4Y is______ .
From the polar coordinate equation of curve C: ρ 2-2 ρ cos θ - 4 ρ sin θ + 4 = 0, it can be changed into rectangular coordinate equation: x2 + y2-2x-4y + 4 = 0, it can be changed into (x-1) 2 + (Y-2) 2 = 1. The center of circle C (1,2) and radius r = 1 can be obtained. Let 3x + 4Y = t, ∵ point P (x, y) be any point on curve C, and the distance from center of circle C (1,2) to straight line D ≤ R. | 3 + 4 × 2 − t | 32 + 42 ≤ 1, it can be changed into | T-11 | ≤ 5, and the solution is 6 ≤ t ≤ 16 The value range of 3x + 4Y is [6, 16]. So the answer is: [6, 16]
What is the chord length π (SIN) = 4
Method 1: substituting ρ sin (θ + π / 4) = 2 and circle ρ = 4, sin (θ + π / 4) = 1 / 2, θ + π / 4 = π / 6 or 5 π / 6, θ 1 = - π / 12 or θ 2 = 7 π / 12, and θ 2 - θ 1 = 2 π / 3, (θ 2 - θ 1) / 2 = π / 3, then the chord length is 2 * 4 * sin (π / 3) = 4 * root sign 3. Method 2: using the geometric meaning of polar coordinate equation
It takes Xiaohong 10 days and Xiaoli 8 days to read the same book. What is the ratio of reading time and speed between Xiaohong and Xiaoli?
Time: Xiaohong: Xiaoli = 10:8 = 5:4
Speed: Xiaohong: Xiaoli = 1 / 10:1 / 8 = 4:5
How many kilowatts of electricity does a 100 watt light consume in an hour?
RT
One kilowatt hour is one kilowatt hour
100 / 1000 kW * 1 hour = 0.1 kwh, i.e. 0.1 kwh
How many microamperes is 1000 amperes
1000000000 microamperes (1 a = 1000 Ma = 1000000 microamperes)
A 50 meter long fence is used to form a rectangular vegetable garden with one side against the wall. When you ask what is the length and width of the rectangle, what is the maximum area of the garden?
Suppose the area of the garden is s, the side perpendicular to the wall is x, and the side parallel to the wall is 50-2x
S=x（50-2x）
=-2x+50x
=-2（x-25/2）²+625/2
Because a = - 2
S=(50-X)/2*X
Finding the maximum of S
x=25
s=25*12.5=312.5
The length and width are 5cm respectively
When the length and width are equal, the area is the largest.
Side length: 50 △ 3 = 16 and 2 / 3 (m)
Maximum: 16.2/3 × 16.2/3 = 277.7/9 (M2)
When the length is 25 and the width is 12.5, the maximum is 312.5
forehead
So simple
50 △ 4 = 12.5m
12.5 × 12.5 = 156.25 square meters
A: the length is 12.5m, the width is 12.5m, and the maximum value is 156.25m
Xiaoli takes 1 / 4 more road than Xiaohong, and Xiaoli takes 10 / 11 of Xiaohong's time. So, the speed ratio between Xiaoli and Xiaohong is (--)
Xiaoli's journey is one fourth more than Xiaohong's
（1＋1/4）：1＝5：4
The time for Xiaoli to walk is 10 / 11 of Xiaohong's
10/11：1＝10：11
The speed ratio of Xiaoli and Xiaohong is:
5/10：4/11＝11：8
11/8
A 100 watt lamp lights for hours
One degree = 1000 watt hours
1000 watt hours / 100 watt = 10 (hours)
1 degree = 1000 W x H
How many amperes is 100 microamperes?
1 a = 1 000 Ma = 1 000 Ma
1000000 microamps = 1 Amp
100 μ a = 0.0001 a
1 * 10 - 4 power ampere

RELATED INFORMATIONS