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Pre-Calc Homework Solutions 460

# Pre-Calc Homework Solutions 460 - 460 Appendix A5.2 p and...

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39. Let P 1 ( 2 p , y 1 ) be any point on x 5 2 p , and let P ( x , y ) be a point where a tangent intersects y 2 5 4 px . Now y 2 5 4 px 2 y } d d y x } 5 4 p } d d y x } 5 } 2 y p } ; then the slope of a tangent line from P 1 is } x y 2 2 ( 2 y 1 p ) } 5 } d d y x } 5 } 2 y p } y 2 2 yy 1 5 2 px 1 2 p 2 . Since x 5 } 4 y p 2 } , we have y 2 2 yy 1 5 2 p 1 } 4 y p 2 } 2 1 2 p 2 y 2 2 yy 1 5 } 1 2 } y 2 1 2 p 2 } 1 2 } y 2 2 yy 1 2 2 p 2 5 0 y 5 5 y 1 6 ˇ y 1 w 2 w 1 w 4 w p w 2 w . Therefore the slopes of the two tangents from P 1 are m 1 5 and m 2 5 m 1 m 2 5 } y 1 2 2 ( 4 y 1 p 2 2 1 4 p 2 ) } 5 2 1 the lines are perpendicular. 40. Let y 5 ! 1 § 2 § } x 4 § 2 } § on the interval 0 # x # 2. The area of the inscribed rectangle is given by A ( x ) 5 2 x 1 2 ! 1 § 2 § } x 4 § 2 } § 2 5 4 x ! 1 § 2 § } x 4 § 2 } § (since the length is 2 x and the height is 2 y ) A 9 ( x ) 5 4 ! 1 § 2 § } x 4 § 2 } § 2 . Thus A 9 ( x ) 5 0 4 ! 1 § 2 § } x 4 § 2 } § 2 5 0 4 1 1 2 } x 4 2 } 2 2 x 2 5 0 x 2 5 2 x 5 ˇ 2 w (only the positive square root lies in the interval). Since
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