M (- 3,5) n (2,15) on L: 3x-4y + 4 = 0, find point P is the minimum length of PM + PN, find the minimum value of point P coordinates and PM + PN! Before 10:00,

M (- 3,5) n (2,15) on L: 3x-4y + 4 = 0, find point P is the minimum length of PM + PN, find the minimum value of point P coordinates and PM + PN! Before 10:00,

Make the symmetric point Q of m with respect to line L, connect the intersection line L of NQ with point P, and point P is the obtained point
From m (- 3,5) and l: 3x-4y + 4 = 0, let K be the slope of MQ, then k = - 4 / 3. The equation of MQ is Y-5 = - 4 / 3 (x + 3), y = - 4 / 3x + 1. Combining the two equations, the intersection point is (0,1)
Using the midpoint coordinate formula to find the Q coordinate, there are 0 = (- 3 + XQ) / 2,1 = (5 + YQ) / 2, the solution is obtained
XQ = 3, YQ = - 3. It can be seen from the drawing that PM + PN = PQ + PN = QN. It can be obtained from n (2,15) and Q (3, - 3)
QN=√{(2-3)^2+[15-(-3)]^2}=5√13.

Given two points m (- 5,0), n (5,0), if there is a point P on the line such that [PM] - [PN] = 6, then the line is called B-type line, and the following line is given: 1] y = x + 1 [2] y=
Among them, B-type straight line has?

The title is incomplete and can only provide ideas
The trajectory of point P satisfying | PM | - | PN | = 6 is
The right branch of hyperbola x ^ 2 / 9-y ^ 2 / 16 = 1,
If the right branch of the straight hyperbola x ^ 2 / 9-y ^ 2 / 16 = 1 has a common point,
It's a B-line

It is known that the line L passing through the point (1,0) intersects the ellipse x2 / A2 + Y2 / B2 = 1 (a > b > 0 and A2 + B2 > 1) at P and Q, and the coordinates of the midpoint of PQ are (A2 / 2, B2 / 2)
And the vector op ⊥ the vector OQ (o is the coordinate origin)
(1) solving the equation of line L
(2) verification: 1 / A2 + 1 / B2 is the fixed value

Given that the line x-2y + 2 = 0 passes through a vertex and a focus of the ellipse x2a2 + y2b2 = 1, then the equation of the ellipse is___ The centrifugation is___ ．

The intersection of x-2y + 2 = 0 and X axis is a (- 2,0), and the intersection of X axis and Y axis is B (0,1), so a focus of ellipse is f (- 2,0), and a vertex of minor axis is f (0,1), so in ellipse x2a2 + y2b2 = 1 & nbsp; & nbsp; (a ＞ B ＞ 0), C = 2, B = 1,... A = 5