What is the tangent of x^2+xy+2y^2=4,(1,1) ?

The tangent of x^2+xy+2y^2=4,(1,1) is y=-3/5 x+8/5

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tangent of x^2+xy+2y^2=4,(1,1)

tangent of

x^{2} + x y + 2 y^{2} = 4, (1, 1)

y = - \frac{3}{5} x + \frac{8}{5}

Solution steps

Find the slope of

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

Compute the slope of

x^{2} + x y + 2 y^{2} = 4

(1, 1) : m = - \frac{3}{5}

Find the line with slope m=

- \frac{3}{5}

and passing through

(1, 1) : y = - \frac{3}{5} x + \frac{8}{5}

y = - \frac{3}{5} x + \frac{8}{5}

f (x) = 2 x^{6} + 3 x

\frac{5}{e ^{x} + e ^{- x}}

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

f (x) = x^{- \frac{1}{9}}

\frac{( s - 5 )}{( s ^{2} - 6 s + 25 )}

What is the tangent of x^2+xy+2y^2=4,(1,1) ?
The tangent of x^2+xy+2y^2=4,(1,1) is y=-3/5 x+8/5