If the parabola C1 and the parabola C2: y ^ 2 = - 4x are symmetric with respect to the straight line x + y = 2, then the focal coordinate of the parabola C1 is

If the parabola C1 and the parabola C2: y ^ 2 = - 4x are symmetric with respect to the straight line x + y = 2, then the focal coordinate of the parabola C1 is

(2,3), the focal point of the known parabola is (- 1,0), and the symmetry of two parabolas with respect to the straight line is known, that is, the two focal points are symmetrical with respect to the straight line. The symmetry point is (2,3)

Given the circle C1: x ^ 2 + y ^ 2 = 1 and the circle C2: (x-3) ^ 2 + (y-4) ^ 2 = 4, if the tangent Ma and MB from the point m (x, y) to the circle C1 and C2 are equal in length, the trajectory equation of the moving point m is obtained

A (1,1) C1 (0,0) it is easy to find the slope of the straight line C1a = 1. Because l is tangent, it is perpendicular to the radius C1a, so the slope of L = - 1. So the equation of L is y = - x + 2, that is, x + Y-2 = 0 (2). Because C2 is on the straight line y = 2x, we can set the coordinate of C2 as (a, 2a) because it passes through O (0,0), so the radius = √ (a ^ 2 + (2a) ^ 2) = √ 5 * a, let C2

The intersection of ellipse C1: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) and parabola C2: x ^ 2 = 2PY (P > 0) is m. the tangent of parabola C2 at point m passes through the right focus of ellipse

Ellipse C1: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) parabola C2: x ^ 2 = 2PY (P > 0) (1) if M (2, (2 √ 5) / 5), find the standard equation m (2,2 / 5 * √ 5 = 2 / √ 5) of C1 and C2 and substitute it into parabola C2: x ^ 2 = 2PY (P > 0) 2 ^ 2 = 2p * 2 / √ 5p = √ 5 the tangent of parabola C2 at point m dy / DX = D / DX (x ^ 2 / 2P) = x / P