Given that the right angle vertex a of RT triangle ABC moves on the straight line ρ cos θ = 9, C is the origin, and the angle ACB is 30 °, the polar coordinate equation of the trajectory of vertex B is obtained

Given that the right angle vertex a of RT triangle ABC moves on the straight line ρ cos θ = 9, C is the origin, and the angle ACB is 30 °, the polar coordinate equation of the trajectory of vertex B is obtained

The Cartesian coordinates of point a are (9, ρ sin θ)
The following two cases are discussed
If ABC is clockwise, then the rectangular coordinates of point B are (2 ρ / √ 3 * cos (θ - π / 6), 2 ρ / √ 3 * sin (θ - π / 6)), and angle a is a right angle, so the equation AB * AC = 0 can be established, that is, 9 (2 ρ / √ 3 * cos (θ - π / 6) - 9) + ρ sin θ (2 ρ / √ 3 * sin (θ - π / 6) - ρ sin θ) = 0, and ρ ^ 2Sin θ cos θ - ρ (9 √ 3cos θ + 9sin θ) + 81 √ 3 = 0
That is, (ρ sin θ - 9 √ 3) (ρ cos θ - 9) = 0, where ρ cos θ = 9 is the trajectory of a, so the trajectory of B is ρ sin θ = 9 √ 3
If ABC is counter clockwise, then the rectangular coordinates of point B are (2 ρ / √ 3 * cos (θ + π / 6), 2 ρ / √ 3 * sin (θ + π / 6)), and angle a is a right angle, so we can establish the equation AB * AC = 0, that is, 9 (2 ρ / √ 3 * cos (θ + π / 6) - 9) + ρ sin θ (2 ρ / √ 3 * sin (θ + π / 6) - ρ sin θ) = 0, and simplify to ρ ^ 2Sin θ cos θ + ρ (9 √ 3cos θ - 9sin θ) - 81 √ 3 = 0
That is, (ρ sin θ + 9 √ 3) (ρ cos θ - 9) = 0, where ρ cos θ = 9 is the trajectory of a, so the trajectory of B is ρ sin θ = - 9 √ 3
Draw the image of the function y = | 3 ^ x + 1 | and use the image to answer: when k is the value, the equation k = | 3 ^ x + 1 | has no solution
Careless child, the function should be y = | 3 ^ X - 1 | and of course the equation k = | 3 ^ X - 1|
(1)y=3^x ------->(2)y=3^x-1--------->(3)y=|3^x -1|
Translate (1) image down one unit to get (2)
(2) The asymptote of image passing (0,0) is y = - 1
Keep the (2) image above the x-axis, and clip it between the x-axis and y = - 1
This part of the image is sandwiched between the x-axis and the x-axis
The image of (3) y = | 3 ^ X - 1 | is obtained by taking y = 1 as the asymptote between y = 1 and y = 1
The number of solutions of equation k = | 3 ^ X - 1 | is the image of y = K and y = | 3 ^ X - 1 | respectively
The number of intersections
When k
Given that the two ends of the line AB with length a move on the positive half axes of X and Y respectively, the trajectory equation of the vertex C of the equilateral triangle ABC (C, O in x) is obtained
Please use the knowledge of parametric equation to solve the problem. Given that the two ends of the line AB with length a move on the positive half axis X and Y respectively, find the trajectory equation of the vertex C of the regular triangle ABC (C and o are on both sides of AB)
It's too busy. You can make the equation slowly according to the meaning of the question
Draw the graph of function y = absolute value (3 ^ x-1), and find out when k is the value, the function has no solution? One solution? Two solutions?
When using the image to answer what is the value of K, the absolute value of the equation (3 ^ x-1) = k has no solution? One solution? Two solutions?
It can be seen from the graph (only x ∈ [- 1,1]), when x ∈ (- ∞, 1], y ∈ [0,1], when x ∈ [1, ∞), y ∈ [0, ∞). Therefore, for the equation y (x) = ABS ((3 ^ x) - 1) = k: 1), when K & lt; 0, there is no solution; 2) when k = 0 or K ≥ 1, there is a unique solution; 3) when k ∈ (0,1), there are two solutions
In the plane rectangular coordinate system, the coordinates of a, B and C are a (5,0), B (0,3) and C (5,3), respectively. O is the origin of the coordinates, and point E is on the line BC. If △ AEO is an isosceles triangle, the coordinates of point E can be obtained. (draw the image, no need to write the calculation process.)
The graph is as follows: (1) if isosceles △ AEO takes a as vertex, then E (1,3); (2) if isosceles △ AEO takes e as vertex, then E (2.5,3); (3) if isosceles △ AEO takes O as vertex, then E (4,3)
If f (x-1) + F (x-3) does not satisfy the absolute value of all functions X-1,
If f (x) satisfies f (x-3 / 1-x) + F (3 + X / 1-x) = x for all real numbers x whose absolute value x is not equal to 1
Wrong number. There's no equation
In the plane rectangular coordinate system, the coordinates of a, B and C are a (5,0), B (0,3) and C (5,3), respectively. O is the coordinate origin, if △ AEO is isosceles
Triangle. If point E is on line BC, the coordinates of point E can be obtained. If point E is on line BC, the coordinates of point E can be obtained
(1) If on the BC line segment, there are three cases EO = EA: then it is obvious that e is at the midpoint of BC, and the coordinates of E are (2.5, 3); ①
OA = OE, then OE = 5, ∵ ob = 3, ∵ be = 4, namely e (4,3); ②
Ao = AE, similarly, we get e (1,3); ③
(2) If on the BC line, in addition to the above cases, there are cases where ∠ e0a and ∠ EAO are obtuse angles. E in (2) can be obtained by Y-axis symmetry
And (3) e (9,3), which is symmetrical with a straight line x = 5, is also an isosceles triangle
The function f (x) = 2|x − 4| (x ≠ 4) a & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; (x = 4), if the function y = f (x) - 2 has three zeros, then the value of real number a is ()
A. -4B. -2C. 2D. 4
Function f (x) = 2|x − 4| (x ≠ 4) a & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; (x = 4) then function y = f (x) - 2 = 2|x − 4 | 2 & nbsp; & nbsp; & nbsp; (x ≠ 4) a = 2 & nbsp; & nbsp; & nbsp; & nbsp; (x = 4) if x ≠ 4, then 2 | x − 4 | 2 = 0, then x = 3 or x = 5
As shown in the figure, in the plane rectangular coordinate system, O is the origin of the coordinate, and the coordinates of ABC three points are a (0,8), B (10,0) C (7,4) ad ‖ X axis, and the line BC
Intersection point D, moving point P starts from a and moves along the line of broken line AOB at a speed of one unit per second. The moving time is T seconds. Let the area of △ ACP be s
What is the value of T when △ PBC is an isosceles triangle?
If △ PBC is an isosceles triangle, point P should be on ob, so let P (x, 0)
(1) When the two waists of the isosceles triangle △ PBC are PC = BC = 5, there is (7-x) ^ 2 + 16 = 25, and the solution is x = 4, then t = 8 + 4 = 12;
(2) When the two waists of the isosceles triangle △ PBC are Pb = BC = 5, there is 10-x = 5, and the solution is x = 5, then t = 8 + 5 = 13;
(3) When the two waists of the isosceles triangle △ PBC are Pb = PC, (7-x) ^ 2 + 16 = (10-x) ^ 2, the solution is x = 35 / 6, then t = 8 + 35 / 6 = 83 / 6
When t = 12, PC = BC; when t = 13, Pb = BC; when t = 83 / 6, PC = Pb
Let f (x) = x2 + 2x (x ≠ 0), when a > 1, the number of real roots of the equation f (x) = f (a) is______ .
F ′ (x) = 2x − 2x2. ① when x ＞ 1, that is, f ′ (x) ＞ 0, the function f (x) = x2 + 2x is an increasing function on (1, + ∞). ② when 0 ＜ x ＜ 1, f ′ (x) ＜ 0, the function f (x) = x2 + 2x is a decreasing function on (0, 1). ③ when x ＜ 0, f ′ (x) ＜ 0, the function f (x) = x2 + 2x is a decreasing function on (- ∞, 0). The diagram is as follows: then the equation f (x) = f (a) So the answer is: 3