What is the tangent of 2x^3+2y^3-9xy=0,(2,1) ?

The tangent of 2x^3+2y^3-9xy=0,(2,1) is y= 5/4 x-3/2

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tangent of 2x^3+2y^3-9xy=0,(2,1)

tangent of

2 x^{3} + 2 y^{3} - 9 x y = 0, (2, 1)

y = \frac{5}{4} x - \frac{3}{2}

Solution steps

Find the slope of

2 x^{3} + 2 y^{3} - 9 x y = 0 : \frac{d y}{d x} = \frac{- 2 x ^{2} + 3 y}{2 y ^{2} - 3 x}

Compute the slope of

2 x^{3} + 2 y^{3} - 9 x y = 0

(2, 1) : m = \frac{5}{4}

Find the line with slope m=

\frac{5}{4}

and passing through

(2, 1) : y = \frac{5}{4} x - \frac{3}{2}

y = \frac{5}{4} x - \frac{3}{2}

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What is the tangent of 2x^3+2y^3-9xy=0,(2,1) ?
The tangent of 2x^3+2y^3-9xy=0,(2,1) is y= 5/4 x-3/2