The ellipse is x ^ 2 / 25 + y ^ 2 / 9 = 1, point a is (4,0) and B is (3,1). Find a point P on the ellipse to minimize | Pb | + 5 / 4 | PA |

The ellipse is x ^ 2 / 25 + y ^ 2 / 9 = 1, point a is (4,0) and B is (3,1). Find a point P on the ellipse to minimize | Pb | + 5 / 4 | PA |

The eccentricity of ellipse is 4 / 5, that is to say, the distance from point P to focus a | PA | is equal to the distance from P to directrix = 25 / 4, and point B is in the circle, so the minimum value of | Pb | + 5 / 4 | PA | is the distance from point B to directrix = 25 / 4, that is, 25 / 4-3 = 13 / 4

There are two points a (2,2) B (3,0) in the ellipse x ^ 2 / 25 + y ^ 2 / 16 = 1. P is any point of the ellipse. If we want to minimize | PA | + 5 / 3 | Pb |, we can find that | PA | + 5 / 3 | Pb |,
Then the minimum value is?

When doing the question, first write out the parameters of the ellipse, and then analyze the mystery of the coefficient 5 / 3. If it is a fill in question or a short answer question, then directly write out the distance from the right guide line to point A. in order to consolidate our knowledge, we also find out the specific coordinates of the best position of point P and Q

Given that the point P (x, y) is a moving point on the circle (x-3) 2 + (Y-3) 2 = 6, then the maximum value of YX is______ ．

Let YX = k, then y = KX, when the line y = KX is tangent to the circle (x-3) 2 + (Y-3) 2 = 6, K has the maximum value, that is: | 3K − 3| 1 + K2 = 6, the solution is 3 ± 2, so the maximum value of YX is 3 + 2, so the answer is: 3 + 2