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

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

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

tangent of

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

y = - x + 2

Solution steps

Find the slope of

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

Compute the slope of

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

(1, 1) : m = - 1

Find the line with slope m=

- 1

and passing through

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

y = - x + 2

x \to 7 lim (\frac{( x ^{2} - 5 x - 14 )}{x - 7})

y^{^{'}} + \frac{1}{200 + t} y = 0, y (0) = 60

\int \frac{4 x ^{2} sin ( 2 x )}{9} d x

n = 2 \sum \infty \frac{n - 2}{n 5 ^{n}}

n = 3 \sum \infty \frac{7 ^{n}}{1 4 ^{n}}

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