What is the tangent of 6x^2+xy+6y^2=13,(1,1) ?

The tangent of 6x^2+xy+6y^2=13,(1,1) is y=-x+2

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tangent of 6x^2+xy+6y^2=13,(1,1)

tangent of

6 x^{2} + x y + 6 y^{2} = 13, (1, 1)

y = - x + 2

Solution steps

Find the slope of

6 x^{2} + x y + 6 y^{2} = 13 : \frac{d y}{d x} = \frac{- 12 x - y}{x + 12 y}

Compute the slope of

6 x^{2} + x y + 6 y^{2} = 13

(1, 1) : m = - 1

Find the line with slope m=

- 1

and passing through

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

y = - x + 2

y^{^{''}} + 9 y = 5 csc^{2} (3 t)

x \to 0 lim (\frac{1}{2 + 3 ^{\frac{1}{x}}})

n = 1 \sum \infty \frac{1}{( n + 1 )}

What is the tangent of 6x^2+xy+6y^2=13,(1,1) ?
The tangent of 6x^2+xy+6y^2=13,(1,1) is y=-x+2