In the polar coordinate system, the polar coordinate equation of the line passing through the center of the circle P = 6cosa and perpendicular to the polar axis is?

p=6cosa
To rectangular coordinate equation
p^2=6pcosa
x^2+y^2=6x
(x-3)^2+y^2=9
Center (3,0)
A line perpendicular to the X axis
x=3
To polar coordinate equation
pcosa=3

In the polar coordinate system, the polar coordinate equation of the line passing through the center of the circle ρ = 6cos θ and perpendicular to the polar axis is

The original polar coordinates can be changed to x ^ 2 + y ^ 2-6cosx = 0 (multiply both sides by P)
So it's equivalent to a straight line of (3,0)
Then I'll draw another triangle to confirm the triangle relationship
That's 3 = PCOS theta

If point P is on the plane region 2x − y + 2 ≥ 0x + y − 2 ≤ 02y − 1 ≥ 0 and point q is on the curve x2 + (y + 2) 2 = 1, then the minimum value of | PQ | is______ ．

To make the feasible region of | PQ | as shown in the figure, as long as the distance from the center of circle C (0, - 2) to p is the minimum, combined with the graph, when p is at point (0, 12), the minimum | CP | is 12 + 2 = 52, and because the radius of circle is 1, the minimum | PQ | is 32, so the answer is: 32

Let p be on the curve y = 1 / 2E ^ X and Q be on the curve y = ln (2x), then the minimum value of ∣ PQ ∣ is
thank you

&That's the answer

As shown in the figure, given the line segments a, B, C, use a compass and ruler to make the line segments, so that they are equal to a + 2b-c. A is the longest, B is the shortest, C is shorter than a, and longer than B
As shown in the figure, given the line segments a, B and C, use a compass and a ruler to make the line segments so that they are equal to a + 2b-c. A is the longest and B is the shortest. No C is shorter than a.

1> Fix a point. The first side of the ruler is close to this point. Fix the ruler and don't move
2> Compare the length of a with a compass and make a point on the side of the ruler against the fixed point
3> Compare the length of B with a compass and continue to make 2 points after the 2 > point
4> Compare the length of C with the compass, continue to hit the last point of 3 > and hit one point back
Then draw a line segment between the 1 > fixed point and the 4 > fixed point, and you are done

It is known that one root of the equation x & # 178; + kx-2 = 0 is the same as the root of the equation x + 1 / X-1 = 3, and the value of K is the other root of the equation x & # 178; + kx-2 = 0

I learned the second volume of junior high school

We know the equation x & # 178; - 2 (K-3) x + K & # 178; - 4k-1 = 0 about X. if the two roots of the equation are taken as abscissa and the point of ordinate is just on the image of inverse scale function y = m / x, then the minimum value of M satisfying the condition is?

The equation has roots
Then 4 (K-3) & # 178; - 4 (K & # 178; - 4k-1) ≥ 0
That is - 8K + 40 ≥ 0
K ≤ 5
x1x2=k²-4k-1
x2=m/x1
That is, M = x1x2
=k²-4k-1
=(k-2)²-5
That is, when k = 2, the minimum value of M is - 5
In conclusion, the minimum value of M satisfying the condition is - 5

The equation x & # 178; - 2 (K-3) + K & # 178; - 4k-1 = 0, has a root of 1, and finds the value of K

x=1
Substituting
1-2k+6+k²-4k-1=0
k²-6k+6=0
k=3±√3

In the polar coordinate system, the polar coordinate equation of the line passing through the center of the circle P = 6cosa and perpendicular to the polar axis is?