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Exercise 9.8 · Q9

Q.The curve y=(x−2)2+1y=(x-2)^2+1 has a minimum point at PP. A point QQ on the curve is such that the slope of PQPQ is 2. Find the area bounded by the curve and the chord PQPQ.

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Find the vertex, solve the slope condition for QQ, confirm the chord is above the curve, and integrate (chord −- curve).

Step 1. Locate PP. y=(x−2)2+1≥1y=(x-2)^2+1\ge1 for all xx, with equality at x=2x=2; so P=(2,1)P=(2,1).

Step 2. Locate QQ. For a point Q=(x,(x−2)2+1)Q=(x,(x-2)^2+1) on the curve, slope of PQPQ is (x−2)2+1−1x−2=x−2\dfrac{(x-2)^2+1-1}{x-2}=x-2 (for x≠2x\ne2). Setting this to 22: x−2=2⇒x=4x-2=2\Rightarrow x=4, so Q=(4,(4−2)2+1)=(4,5)Q=(4,(4-2)^2+1)=(4,5).

Step 3. Equation of chord PQPQ. Slope 22 through (2,1)(2,1): y−1=2(x−2)⇒y=2x−3y-1=2(x-2)\Rightarrow y=2x-3.

Step 4. Identify which is on top. At x=3x=3 (between 22 and 44): chord =2(3)−3=3=2(3)-3=3; curve =(3−2)2+1=2=(3-2)^2+1=2. The chord lies above the curve on [2,4][2,4]. …

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