Given the line 4x + 3Y + 6 = 0 and x = - 1, find the minimum of the sum of the distances between the moving point P on y ^ 2 = 4x and the two lines

Given the line 4x + 3Y + 6 = 0 and x = - 1, find the minimum of the sum of the distances between the moving point P on y ^ 2 = 4x and the two lines

If x = - 1 is a guide line, then p to x = - 1 equals PF
F is the focus (1,0)
The intersection of the vertical line 4x + 3Y + 6 = 0 and the parabola is p
So the distance and the minimum is the distance from F to the straight line
So the minimum value = | 4 + 0 + 6 | / √ (4 & sup2; + 3 & sup2;) = 2