What is the tangent of 3x^2+xy+3y^2=7,(1,1) ?

The tangent of 3x^2+xy+3y^2=7,(1,1) is y=-x+2

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tangent of 3x^2+xy+3y^2=7,(1,1)

tangent of

3 x^{2} + x y + 3 y^{2} = 7, (1, 1)

y = - x + 2

Solution steps

Find the slope of

3 x^{2} + x y + 3 y^{2} = 7 : \frac{d y}{d x} = \frac{- 6 x - y}{x + 6 y}

Compute the slope of

3 x^{2} + x y + 3 y^{2} = 7

(1, 1) : m = - 1

Find the line with slope m=

- 1

and passing through

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

y = - x + 2

\frac{sin ( t )}{2 + 2 cos ( t )}

\frac{1}{( s + 1 ) ^{2} + 4}

\int \frac{x}{x + 3} d x

\frac{d}{d x} (\frac{d y}{d x}) - y = 0

\int_{0}^{N} e^{- s t} e^{2 t} d t

What is the tangent of 3x^2+xy+3y^2=7,(1,1) ?
The tangent of 3x^2+xy+3y^2=7,(1,1) is y=-x+2