What is the tangent of 8x^2+xy+8y^2=17,(1,1) ?

The tangent of 8x^2+xy+8y^2=17,(1,1) is y=-x+2

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tangent of 8x^2+xy+8y^2=17,(1,1)

tangent of

8 x^{2} + x y + 8 y^{2} = 17, (1, 1)

y = - x + 2

Solution steps

Find the slope of

8 x^{2} + x y + 8 y^{2} = 17 : \frac{d y}{d x} = \frac{- 16 x - y}{x + 16 y}

Compute the slope of

8 x^{2} + x y + 8 y^{2} = 17

(1, 1) : m = - 1

Find the line with slope m=

- 1

and passing through

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

y = - x + 2

y = 4 x^{3} - 12 x + 3

\int x^{2} - 5 x + 2 d x

(- 2.7) (- 3.4)

\int \frac{14}{( t ^{2} + 1 ) ^{\frac{3}{2}}} d t

\frac{d}{d x} (2^{3 x^{4} - 3 x})

What is the tangent of 8x^2+xy+8y^2=17,(1,1) ?
The tangent of 8x^2+xy+8y^2=17,(1,1) is y=-x+2