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

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saj-94 | (Level 1) Salutatorian

Posted September 4, 2013 at 4:31 PM via web

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A curve is such that dy/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.

 A) find the coordinates of P.

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

Posted September 4, 2013 at 4:54 PM (Answer #1)

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`dy/dx = 5-8/x^2`

`y = int (5-8/x^2) dx`

`y = 5x-8/x+C`

`3y+x = 17`

`y = (-1/3)x+17/3`

A line perpendicular to this has gradient as `-1/((-1/3)) = 3`

If line is perpendicular to P then the tangent at P is perpendicular to line.

So the tangent of the curve at P has gradient 3.

The tangent at any point is given by `dy/dx` .

`dy/dx = 5-8/x^2`

`3 = 5-8/x^2`

`x^2 = 4`

`x = +-2`

It is given that x-coordinate is positive. So x = 2

Then `y = (-1/3)xx2+17/3 = 5`

So the coordinates of P is (2,5)

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