Proof: the distance from any point on an equiaxed hyperbola to the center of symmetry is the equal proportion median of the distance from him to two focal points

Proof: the distance from any point on an equiaxed hyperbola to the center of symmetry is the equal proportion median of the distance from him to two focal points

Suppose that the hyperbola is x ^ 2-y ^ 2 = a ^ 2, then we can know the eccentricity of the hyperbola e e = √ 2. For the convenience of research, we can set a point P (x0, Y0) in the right branch of the hyperbola, and in the first quadrant. The symmetry center of the hyperbola is o point, and the left and right focus of the hyperbola are F1 (- C, 0), F2 (C, 0), then the vector Pf1 = (- c-x0, - Y0), the vector PF2 = (c-x0, - Y0), then the vector Pf1 * vector PF2 = x0 ^ 2 + Y0 ^ 2-C ^ 2, From hyperbolic equation, we can get x0 ^ 2 + Y0 ^ 2-C ^ 2 = 2x0 ^ 2-3a ^ 2. Focal radius Pf1 = ex0 + A, focal radius PF2 = ex0-a, points F1, O and F2 are collinear, and O is the midpoint of line F1F2, so we have vector Po = 1 / 2 * (vector Pf1 + vector PF2). After doing these preparations, we can start to solve the problem
(line PO) ^ 2 = (vector PO) ^ 2 = 1 / 4 * (vector Pf1 + vector PF2) ^ 2 = 1 / 4 [(vector Pf1) ^ 2 + (vector PF2) ^ 2 + 2 (vector Pf1) * (vector PF2)] = 1 / 4 [(ex0 + a) ^ 2 + (ex0-a) ^ 2 + 2x0 ^ 2-3a ^ 2] = 1 / 4 (8x0 ^ 2-4a ^ 2) = 2x0 ^ 2-A ^ 2 = (√ 2x0 + a) (√ 2x0-a) = (ex0 + a) (ex0-a) = focal radius Pf1 * focal radius PF2
This method belongs to vector method, and the process is not difficult. It mainly uses a conclusion of vector, the focal radius of hyperbola, and the eccentricity of equiaxed hyperbola √ 2 to solve the problem. It may be a bit complicated, but the idea is very clear, so you should be able to understand it