What is the tangent of 2x^2+xy+2y^2=5,(1,1) ?

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

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tangent of 2x^2+xy+2y^2=5,(1,1)

tangent of

2 x^{2} + x y + 2 y^{2} = 5, (1, 1)

y = - x + 2

Solution steps

Find the slope of

2 x^{2} + x y + 2 y^{2} = 5 : \frac{d y}{d x} = \frac{- 4 x - y}{x + 4 y}

Compute the slope of

2 x^{2} + x y + 2 y^{2} = 5

(1, 1) : m = - 1

Find the line with slope m=

- 1

and passing through

(1, 1) : y = - x + 2

y = - x + 2

\int x (x^{2} + 4)^{2} d x

\int \frac{sec ^{4} ( t )}{tan ^{2} ( t )} d t

\int \frac{tan ( ln ( 5 x + 6 ) )}{5 x + 6} d x

\frac{d}{d x} (\frac{x ^{2} + 2 x}{4 - 5 x})

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

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