1Number 1ExplanationPoint (a,b)(a, b)(a,b) on the curve y=x2+2y = x^2 + 2y=x2+2 has the closest distance to the line y=xy = xy=x. The value of a+ba + ba+b that satisfies is....2142\frac{1}{4}2412122\frac{1}{2}2212342\frac{3}{4}2433333143\frac{1}{4}341ExplanationPoint (a,b)(a, b)(a,b) is located on the curve y=x2+2y = x^2 + 2y=x2+2. Substitute aaa and bbb into the curve function to get b=a2+2b = a^2 + 2b=a2+2 The point becomes (a,a2+2)(a, a^2 + 2)(a,a2+2). Then determine the distance from point (a,a2+2)(a, a^2 + 2)(a,a2+2) to the line y=xy = xy=x distance=∣pm+qn+c∣p2+q2\text{distance} = \frac{|pm + qn + c|}{\sqrt{p^2 + q^2}}distance=p2+q2∣pm+qn+c∣f(a)=∣a−(a2+2)∣12+(−1)2f(a) = \frac{|a - (a^2 + 2)|}{\sqrt{1^2 + (-1)^2}}f(a)=12+(−1)2∣a−(a2+2)∣f(a)=a2−a+22f(a) = \frac{a^2 - a + 2}{\sqrt{2}}f(a)=2a2−a+2f′(a)=12(2a−1)f'(a) = \frac{1}{\sqrt{2}}(2a - 1)f′(a)=21(2a−1) The closest distance is the minimum distance. The minimum distance occurs when f′(a)=0f'(a) = 0f′(a)=012(2a−1)=0\frac{1}{\sqrt{2}}(2a - 1) = 021(2a−1)=0a=12a = \frac{1}{2}a=21 The minimum distance occurs when a=12a = \frac{1}{2}a=21, substitute to get the value of bbb b=a2+2b = a^2 + 2b=a2+2b=(12)2+2b = \left(\frac{1}{2}\right)^2 + 2b=(21)2+2b=94b = \frac{9}{4}b=49 The sum a+ba + ba+b 12+94=114=234\frac{1}{2} + \frac{9}{4} = \frac{11}{4} = 2\frac{3}{4}21+49=411=243
