What is the tangent of x^2+xy+2y^2=8,(-2,-1) ?

The tangent of x^2+xy+2y^2=8,(-2,-1) is y=-5/6 x-8/3

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tangent of x^2+xy+2y^2=8,(-2,-1)

tangent of

x^{2} + x y + 2 y^{2} = 8, (- 2, - 1)

y = - \frac{5}{6} x - \frac{8}{3}

Solution steps

Find the slope of

x^{2} + x y + 2 y^{2} = 8 : \frac{d y}{d x} = \frac{- 2 x - y}{x + 4 y}

Compute the slope of

x^{2} + x y + 2 y^{2} = 8

(- 2, - 1) : m = - \frac{5}{6}

Find the line with slope m=

- \frac{5}{6}

and passing through

(- 2, - 1) : y = - \frac{5}{6} x - \frac{8}{3}

y = - \frac{5}{6} x - \frac{8}{3}

\int 4 x - x x d x

\frac{d}{d x} (\frac{9}{4 - x})

f (x) = 3 x^{3}, at x = 3

\frac{\partial}{\partial x} (x^{2} e^{y} + x - 2 y)

\int \frac{x ^{2} + 2 x + 1}{x ^{3} + x} d x

What is the tangent of x^2+xy+2y^2=8,(-2,-1) ?
The tangent of x^2+xy+2y^2=8,(-2,-1) is y=-5/6 x-8/3