If point P is on ellipse x 216 + y 29 = 1, find the maximum distance and minimum distance from point P to line 3x-4y = 24

If point P is on ellipse x 216 + y 29 = 1, find the maximum distance and minimum distance from point P to line 3x-4y = 24

Since the point P is on the ellipse x216 + Y29 = 1, let P (4cos θ, 3sin θ), then d = | 12cos θ − 12sin θ − 24| 5, that is, d = | 122cos (θ + π 4) − 24| 5, when cos (θ + π 4) = 1, Dmax = 125 (2 + 2); when cos (θ + π 4) = 1, Dmin = 125 (2 − 2)
Cube of a + square of 4A + 4A
a³+4a²+4a
=a(a²+4a+4)
=a(a+2)²
a(a+2)²
1. 8 ^ n △ 4 ^ n = division of the same base power)
8^n÷4^n=（2³）^÷（2²）^n=2^n
8^n÷4^n=4^n
Same base power multiplication and division, base multiplication and division, index unchanged
8^n÷4^n
= 2^(3n) / 2^(2n)
=2^(3n-2n)
=2^n
If the line 3x + 4Y + 2m = 0 intersects the circle x ^ 2 + y ^ 2-4x = O at two points a and B, and OA is perpendicular to ob, (o is the circle point), then M =?
Please try to have a process
Have you studied parametric equations?
Let the coordinates of points on a circle be (2 + 2Sin θ, cos θ)
One point is (2 + 2Sin θ, cos θ) and the other is (2 + 2 (sin θ + 90), cos (θ + 90)) to bring these two coordinates into the linear equation and then eliminate the parameters
How much is a cube - 4A square + 4A?
A cube-4a square + 4A = a (A-2) & #178;
A (the square of a-4a + 4) = the square of a (A-2)
as follows
There are about 150 billion stars in the Milky way. The distance between the nearest star (nearby Centaurus) and the sun is about 4.22 light years (1 light year in a year, 1 light year ≈ 9.45 × 10 12 power). If the speed of a plane is 350 meters per second, how many decimals are needed?
1 light year ≈ 9.45 × 10 to the 12th power
This unit is kilometers
So 1 light year is about 9.45 × 10 15 cubic meters
So the time is 4.22 × 9.45 × 10 to the 15th power △ 350 seconds
1 year = 365 × 24 × 3600 seconds
So the time is 4.22 × 9.45 × 10 to the 15th power △ 350 (365 × 24 × 3600) = 3.6 × 10 to the 6th power
So it's 3.6 × 10 years to the sixth power
The time is 4.22 × 9.45 × 10 to the 15th power △ 350 (365 × 24 × 3600) = 3.6 × 10 to the 6th power
So it's 3.6 × 10 years to the sixth power
Because the distance between the earth and the sun is about 8 minutes, compared with the distance between the sun and the nearby star of Sagittarius, it can be ignored. So the result is 4.22 * 9.45 * 10 ^ 12 / 350 / 3600 / 24 / 365 = 3613.0 years = 3.6x10 ^ 3 years
4.22 × 9.45 × 10 ^ 12 / 350 (365 × 3600) = 3.6 × 10 ^ 6 years
The distance is 4.22 light years, the propagation speed of light is 3 × 10 8 Power M / s, and the time of a year is 365 × 24 × 3600s in seconds, so the nearest star is 4.22 × 3 × 10 8 Power M / s × 365 × 24 × 3600s = 39924576 × 10 9 power m / s
Or according to the tip of the title: 1 light year ≈ 9.45 × 10 12 power kilometers, then the nearest star distance from the sun is: 9.45 × 10 12 power kilometers × 4.22 = 39.879 × 10 12 power kilometers = 3.987
The distance is 4.22 light years, the propagation speed of light is 3 × 10 8 Power M / s, and the time of a year is 365 × 24 × 3600s in seconds, so the nearest star is 4.22 × 3 × 10 8 Power M / s × 365 × 24 × 3600s = 39924576 × 10 9 power m / s
Or according to the tip of the title: 1 light-year ≈ 9.45 × 10 12 power kilometers, then the nearest star distance from the sun is: 9.45 × 10 12 power kilometers × 4.22 = 39.879 × 10 12 power kilometers = 3.9879 × 10 16 power meters (note that "kilometers" have been installed as "meters")
When the speed of the supersonic aircraft is 350 m / s, the time required is: 3.9879 × 10 of the 16th power meter / 350 (s) = 113.94 × 10 of the 12th power second = 3.6 years
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The line 3x + 4Y + M = 0 intersects the circle x ^ 2 + (Y-5 / 2) ^ 2 = 25 / 4 at two points a and B, and OA ⊥ ob is used to find the value of M
It can be seen that the radius of the circle is 5 / 2, and the center of the circle is (0,5 / 2), so the circle passes through point o,
From OA perpendicular to ob, AB is the diameter, so the line 3x + 4Y + M = 0 passes through the center of the circle (0,5 / 2)
The coordinate (0,5 / 2) is brought into the linear equation,
4*5/2+m=0
m=-10
Cube of a - 4A = a (a + 2) * ()
Cube of a - 4A
=a(a²-4）
=a(a+2)(a-2)
1. If X-2 / 1 + (2Y + 1) to the second power = 0, then (x to the second power) + (y to the second power) = ()
2. The original number of approximate number 1.8 is not less than () and less than ()
3. If a = 5, B = 2 and a + B = B + A, then B-A = ()
1. If X-2 / 1 + (2Y + 1) to the second power = 0, then (x to the second power) + (y to the second power) = ()
Because the power of 1 / 2 x + (2Y + 1) is 0
So X-1 / 2 = 0, then x = 1 / 2
The second power of (2Y + 1) = 0, 1 / 2 of y = -
So (x to the second power) + (y to the second power) = 1 / 4 + 1 / 4 = 1 / 2
2. The original number of approximate number 1.8 is not less than (1.75) and less than (1.85)
3. If a = 5, B = 2 and a + B = B + A, then B-A = (- 3, - 7)
1. If X-2 / 1 + (2Y + 1) to the second power = 0, then (x to the second power) + (y to the second power) = (1 / 2)
2. The original number of approximate number 1.8 is not less than (1.75) and less than (1.85)
3. If a = 5, B = 2 and a + B = B + A, then B-A=
(-3 ,-7)
1.1/2
2.1.75 1.85
3 a must be positive. To make B + a greater than 0.
So the answer is - 3 or - 7
Given that the line y = ax + 1 and hyperbola 3x ^ 2 - y ^ 2 = 1 intersect at two points a and B, whether there is a real number a, so that a and B are closed
If it exists, find a; if it does not exist, explain the reason
Y = ax + 1 to 3x ^ 2 - y ^ 2 = 1
We get: (3-A ^ 2) x ^ 2 - 2ax-2 = 0
X1+X2=2a/(3-a^2)
So Y1 + y2 = a (x1 + x2) + 2
If it exists, then [(x1 + x2) / 2 (Y1 + Y2) / 2] must be on the line y = 3x
Take it into account
A=1
So there is