Both parabola and circle are located on the positive semi-axis of X (that is, x≥0) and are symmetrical about X, so only the minimum area of △PBC formed by the intersection of the tangent of the moving point P on the first quadrant parabola and the circle O at point B and point C is needed.
Let the moving point coordinate P(x,
Y), two straight lines tangent to the circle intersect the Y axis at B(0,
B) and C(0,
C) Point, where the straight lines PB and PC are tangent to points A and D respectively, there are ⊙A⊥PB, ⊙D⊥PC, |⊙A|=|⊙D|=|⊙o|= 1.
△⊙oB=△⊙AB,△⊙DP=△⊙AP,△⊙oC=△⊙DC
(All right triangles)
Therefore, the area of △PBC is S=2(△⊙AB area +△⊙AP area +△⊙DC area).
but
2△⊙AP area = |⊙ a ||| AP | = | AP |
2δ⊙AB area = |⊙ A ||| AB | = | AB |
The area of 2 △⊙DC = |⊙ O ||| OC | = | OC |
therefore
S=|AB|+|AP|+|oC|=|PB|+|oC|
You can get the coordinates of tangents A and D, and then get the coordinates of points B and C (related to the coordinates of point P), so as to get the values of |PB| and |oC|, and then get the minimum value of S.
The derivation is complicated, please solve the problem yourself.