What is the tangent of x^2+xy+2y^2=32,(5,1) ?

The tangent of x^2+xy+2y^2=32,(5,1) is y=-11/9 x+64/9

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of x^2+xy+2y^2=32,(5,1)

tangent of

x^{2} + x y + 2 y^{2} = 32, (5, 1)

y = - \frac{11}{9} x + \frac{64}{9}

Solution steps

Find the slope of

x^{2} + x y + 2 y^{2} = 32 : \frac{d y}{d x} = \frac{- 2 x - y}{x + 4 y}

Compute the slope of

x^{2} + x y + 2 y^{2} = 32

(5, 1) : m = - \frac{11}{9}

Find the line with slope m=

- \frac{11}{9}

and passing through

(5, 1) : y = - \frac{11}{9} x + \frac{64}{9}

y = - \frac{11}{9} x + \frac{64}{9}

\frac{d}{d x} (sin^{3} (2 x + 1))

\int_{1}^{3} x^{3} d x

\frac{5}{( s + 4 ) ^{2} ( s + 1 )}

\frac{t}{( t - 1 ) ^{2}}

(1, 1), (- 3, - 3)

What is the tangent of x^2+xy+2y^2=32,(5,1) ?
The tangent of x^2+xy+2y^2=32,(5,1) is y=-11/9 x+64/9