Find the equation of the line tangent to the curve y = 2 - (1/3x) ^ 2 which is perpendicular to the line xy = 0.

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Find the equation of the line tangent to the curve y = 2 - (1/3x) ^ 2 which is perpendicular to the line xy = 0.

Mathematics
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Correcting in the last line. x-y=0.
The product of the slope of the perpendicular lines is equal to minus 1
slope for x-y=0 is equal to 1

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so the slope of the tangent line would be minus 1
Perpendicular slope = - 1 y = 2 - (1/3x)² = 2 - x²/9 ->y' = -2x /9 = -1 => x= 4.5, y = -4.75 Thus Tangent line at ( 4.5, -4.75) : y = -x -.25
Thanks
This is the solution
@viniterranova You're unable to post correctly, are you? Let compare these equations from solution and your post: y = -(1/3)x² + 2 y = 2 - (1/3x) ^ 2
Anyway, thanks for pointing out that it's YOUR posting problem :)

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