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例1 求点 P(4,0)与抛物线 y~2=2x 上的点的距离的最小值。解:设抛物线上一点 Q(x_1,y_1),则y_1~2=2x_1,|PQ|=(x_1-4)~2~(1/2)+y_1~2=(x_1~2-6x_1+16)~(1/2)。∵被开方数二次项的系数为正,∴当 x=3时,(x_1~2-6x_1+16)极小值:=7,|PQ|极小值=7~(1/2)。例2 设 A、B 是椭圆 x~2/a~2+y~2/b~2=1的相邻二顶点,试在(?)上求一点 P,使四边形PAOB 面积为最大。解:设(?)上一点 P(acosθ,bsinθ),则S(?)PAOB=S△AOB+S△PAB
Example 1 Find the minimum value of the distance between the point P (4,0) and the point on the parabola y ~ 2 = 2x. Solution: Let a point on the parabola point Q (x_1, y_1), then y_1 ~ 2 = 2x_1, | PQ | = (x_1-4) ~ 2 ~ (1/2) + y_1 ~ 2 = (x_1 ~ 2-6x_1 + 16 ) ~ (1/2).系 The coefficient of the quadratic term of the square root of the square root is positive, 极 When x = 3, the minimum value of (x_1 ~ 2-6x_1 + 16) = 7, the minimum value of | PQ | 7 ~ (1/2) . Example 2 Let A and B be two adjacent vertices of elliptic x ~ 2 / a ~ 2 + y ~ 2 / b ~ 2 = 1, and try to find a point P on (?) To maximize the area of tetragonal PAOB. Solution: Let (?) Point P (acosθ, bsinθ), then S (?) PAOB = S △ AOB + S △ PAB