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20 July, 10:18

What is the equation of the line tangent to the curve xy=4 at (2,2) ?

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  1. 20 July, 11:45
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    To obtain the slope of this TL at (2,2), differentiate xy=4 with respect to x:

    x (dy/dx) + y (dx/dx) = 0. Then x (dy/dx) = - y, and (dy/dx) = - y/x.

    At (2,2), this slope is - 2/2, or - 1.

    Then the eqn of the TL to the curve xy=4 is y-2 = - 1 (x-2) = - x+2

    Thus, y = 2 - x + 2, or y = - x + 4 (answer)
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