What is the tangent of x^2+xy+2y^2=18,(3,-3) ?

The tangent of x^2+xy+2y^2=18,(3,-3) is y= 1/3 x-4

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tangent of x^2+xy+2y^2=18,(3,-3)

tangent of

x^{2} + x y + 2 y^{2} = 18, (3, - 3)

y = \frac{1}{3} x - 4

Solution steps

Find the slope of

x^{2} + x y + 2 y^{2} = 18 : \frac{d y}{d x} = \frac{- 2 x - y}{x + 4 y}

Compute the slope of

x^{2} + x y + 2 y^{2} = 18

(3, - 3) : m = \frac{1}{3}

Find the line with slope m=

\frac{1}{3}

and passing through

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

y = \frac{1}{3} x - 4

\int \frac{x}{x ^{2} + 6 x + 13} d x

\int x e^{- x} sin (x) d x

x \to - \infty lim (x^{4} e^{x})

\frac{d}{d x} (1 - 2 e^{3 x})

f (x) = (\frac{3 x ^{2}}{2 x + 4})^{3}, x = 2

What is the tangent of x^2+xy+2y^2=18,(3,-3) ?
The tangent of x^2+xy+2y^2=18,(3,-3) is y= 1/3 x-4