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A curve is such thatdy/dx = 5-8/x^2. The line 3y + x = 17 is the normal to the curve at...

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saj-94 | (Level 1) Salutatorian

Posted September 4, 2013 at 4:54 PM via web

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A curve is such thatdy/dx = 5-8/x^2. The line 3y + x = 17 is the normal to the curve at the point P on the
curve. Given that the x-coordinate of P is positive,

B) find the equation of the curve

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted September 4, 2013 at 5:04 PM (Answer #1)

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From part A) we have found that the cordinates of P is (2,5).

http://www.enotes.com/homework-help/curve-such-that-dy-dx-5-8-x-2-line-3y-x-17-normal-451702

`y = 5x-8/x+C` Where C is constant.

This goes through point P. So the curve should satisfy x and y cordinates of P.

P = (2,5)

`5= 5xx2 – 8/2 +C`

`C= -1`

`y=5x-8/x-1`

So the equation of the curve is;

`y=5x-8/x-1`

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted September 4, 2013 at 5:07 PM (Reply #1)

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Please accept the correction for mistype.

`5 = 5xx2-8/2+C`

`5 = 10-4+C`

`C = -1`

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