Given that P is on the curve X = 2 + cos θ, y = sin θ, and point q is on the curve X = T-1, y = root 2T, try to find the minimum value of lpql, and find the coordinates of point Q at this time The root number is 2T. T is also in the root

Given that P is on the curve X = 2 + cos θ, y = sin θ, and point q is on the curve X = T-1, y = root 2T, try to find the minimum value of lpql, and find the coordinates of point Q at this time The root number is 2T. T is also in the root

The curve C at point P is: (X-2) &# 178; + Y & # 178; = 1, which is a circle with (2,0) point as its center and 1 as its radius;
The curve D of Q point is: Y & # 178; = 2T, t = x + 1, that is, Y & # 178; = 2 (x + 1), {y ≥ 0, (∵ t ≥ 0, х x ≥ - 1}. Its image is the upper part of the parabola with the vertex (- 1,0) opening to the right
The combination of the two looks like a fish's head with extra large eyes. Ha ha. It's a joke
This also inspires us: draw a circle with variable radius with (2,0) point as the center, as long as the circle is tangent to the parabola
Let (X-2) ² + Y & #178; = R & #178;, and the parabola y & #178; = 2 (x + 1), eliminate y, make the discriminant equal to zero, find R, subtract the radius of the circle 1 from R, and that is the minimum value of | PQ |. If one of the roots x (> 0) of the quadratic equation of one variable is found, it is the abscissa of point Q
And then, by substituting the parabola and finding y, we get the coordinates of Q
There is a way. I think it's better to leave it to yourself. It's tasteless to eat the steamed bread chewed by others, don't you think?

Let Tan (θ / 2) = t, prove: sin θ = 2T / (1 + T ^ 2), cos θ = (1-T ^ 2) / (1 + T ^ 2), Tan θ = 2T / (1-T ^ 2)
I didn't learn it

In the following, sin and COS refer to sin (θ / 2), cos (θ / 2)
Sin θ = 2sincos / (COS ^ 2 + sin ^ 2) (division of COS ^ 2)
=2t/(1+t^2)
cosθ=(cos^2-sin^2)/(cos^2+sin^2)
=(1-t^2)/(1+t^2)
tanθ=sinθ/cosθ=2t/(1-t^2)
The proof of the universal formula