What is the tangent of x^2+xy+2y^2=16,(-2,-2) ?

The tangent of x^2+xy+2y^2=16,(-2,-2) is y=-3/5 x-16/5

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tangent of x^2+xy+2y^2=16,(-2,-2)

tangent of

x^{2} + x y + 2 y^{2} = 16, (- 2, - 2)

y = - \frac{3}{5} x - \frac{16}{5}

Solution steps

Find the slope of

x^{2} + x y + 2 y^{2} = 16 : \frac{d y}{d x} = \frac{- 2 x - y}{x + 4 y}

Compute the slope of

x^{2} + x y + 2 y^{2} = 16

(- 2, - 2) : m = - \frac{3}{5}

Find the line with slope m=

- \frac{3}{5}

and passing through

(- 2, - 2) : y = - \frac{3}{5} x - \frac{16}{5}

y = - \frac{3}{5} x - \frac{16}{5}

\int (x^{2} - 2 x) (x^{3} - 3 x^{2})^{2} d x

\frac{\partial}{\partial x} ((5 x + 8 y)^{11})

\frac{d ^{2}}{d x ^{2}} (\frac{x ^{2}}{5 + x})

\int_{3}^{6} (2 x - 10) d x

\frac{d}{d x} (- 128 cos (2 x))

What is the tangent of x^2+xy+2y^2=16,(-2,-2) ?
The tangent of x^2+xy+2y^2=16,(-2,-2) is y=-3/5 x-16/5