It is proved that the slope product of AP and BP is colonization

It is proved that the slope product of AP and BP is colonization

First, the rectangular coordinate system is established
Let P (x, y) a (- A, 0) B (a, O)
Then kPa * kbp = Y / (x + a) * y / (x-a) = the square of Y / (the square of x-a)
Let the equation of the ellipse be the square of X divided by the square of a + the square of Y divided by the square of B = 1
If y is represented by x a B, then the square of y = (a times B - B times x) divided by A
By taking the above formula into the first equation, we can get that the slope product is equal to minus B divided by A
I'm learning this right now

Given a (1,0), ellipse C: x ^ 2 / 4 + y ^ 2 / 3 = 1, crossing point a to make a straight line intersection, ellipse C at P and Q, AP = 2qa, then the slope of line PQ

Let P (x1, Y2) Q (X2, Y2). Defined by vector and ellipse, we obtain {(x1-1, Y1) = 2 (1-x2, - Y2), / PA / = 2 / QA /, the equations {x1-1 = 2 (1-x2), 2 - (1 / 2) x1 = 2 (2 - (1 / 2) x2). 2-1 / 2x1 = 2 (2-1 / 2x2) solution, Q {7 / 4, + - (3 √ 5 / 8)}, P {5 / 2, + - (3 √ 3 / 4)} solution, k =

Given point a (1,0), ellipse CX ^ 2 / 4 + y ^ 2 / 3 = 1, cross point a as a straight line, ellipse at P and Q, vector AP = 2, vector QA, then the slope of line PQ is

Let P (x1, Y1), q (X2, Y2) be obtained by vector AP = 2, vector QA, Y1 = - 2Y2, X1 = - 2x2 + 3, and (X2, Y2), (x1, Y1) be combined with ellipse C: x ^ 2 / 4 + y ^ 2 / 3 = 1 respectively, then X1 ^ 2 / 4 + Y1 ^ 2 / 3 = 1, X2 ^ 2 / 4 + Y2 ^ 2 / 3 = 1, and (4x2 ^ 2-12x2 + 9) / 4 + 4y2 ^ 2 / 3 = 1 by replacing X1 and Y1 with x2 and Y2