Through the focus F of parabola y2 = 4ax (a > 0), make two mutually perpendicular focus chords AB and CD, and find the minimum value of | ab | + | CD |

Through the focus F of parabola y2 = 4ax (a > 0), make two mutually perpendicular focus chords AB and CD, and find the minimum value of | ab | + | CD |

The focus f coordinate of the focus of the focus of the parabolic is (a, 0) (a, 0), and the AB equation of the straight line is y = K (x-a), then the CD equation is y = - 1K (x − a), and the CD equation is y = - 1K (x − 1k1k (x-a), the CD equation is y = - 1k1k1k1k (x-a), and the CD equation is y = - 1k1k1k (x-1k (x-a), then the CD equation is y = - 1k1k (x (x − 1K (x − 1K (x (x − a), which is substituted into y2 = 4x2 = 4x2 = 4x2 (x 2 (x 2 = 4x2 (x 2) (2x2 (x 2) (2ak2 + 2 + 2 + 2 + 2 + 4a) to get: k2x2 - (2ak2 - (2ak2 + 2 + 2 + 2 + 4A + 4A + 2 + 4A + 4A + 4A + 2 + 4A + 4A + 2 + 4a) the minimum value of | ab | + | CD | 16A