Find two points of the curve given in a certain plane?

Question

Consider the curve C in the xy-plane given by the equation

x^2y^2-2xy=24.

Find a point a and b on C with a>0 at which the tangent to C has slope m=1.

Find for points a and b

a= 

b=

show work with answer.

Keith · Accepted Answer

Find a point (a , b) on C

otherwise a>0  doesn't make any sense.

x^2 y^2 - 2 x y = 24
. . . . . . implicit derivative
2x y^2 - 2y + dy/dx * (2 x^2 y - 2x) = 0

dy/dx = (2y - 2x y^2) / (2x^2 y - 2x) <== dy/dx = slope

. . . given point has a slope of 1

(2y - 2x y^2) / (2x^2 y - 2x) = 1

(2y - 2x y^2) = (2x^2 y - 2x)

(y - x y^2) = (x^2 y - x)

x + y - x y^2 - x^2 y = 0

x = 1 ; y = -1
or
x = -1 ; y = 1

There are two points were the slope = 1

(1 , -1) and (-1 , 1)

. . . given a>0, (-1 , 1) is disqualified

Slope of C at (1 , -1) = 1

a = 1 ; b = -1

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Find two points of the curve given in a certain plane?

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