What is the tangent of x3+y3-6xy=0 ?

The tangent of x3+y3-6xy=0 is y=-1/((1-2a_{0))^2}x+(2a_{0}^2)/((-2a_{0)+1)^2}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of x3+y3-6xy=0

tangent of

x 3 + y 3 - 6 x y = 0

y = - \frac{1}{( 1 - 2 a _{0} ) ^{2}} x + \frac{2 a _{0}^{2}}{( - 2 a _{0} + 1 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - \frac{a _{0}}{1 - 2 a _{0}})

Find the slope of

x 3 + y 3 - 6 x y = 0 : \frac{d y}{d x} = - \frac{1}{( 1 - 2 x ) ^{2}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - \frac{1}{( 1 - 2 a _{0} ) ^{2}}

Find the line with slope m=

- \frac{1}{( 1 - 2 a _{0} ) ^{2}}

and passing through

(a_{0}, - \frac{a _{0}}{1 - 2 a _{0}}) : y = - \frac{1}{( 1 - 2 a _{0} ) ^{2}} x + \frac{2 a _{0}^{2}}{( - 2 a _{0} + 1 ) ^{2}}

y = - \frac{1}{( 1 - 2 a _{0} ) ^{2}} x + \frac{2 a _{0}^{2}}{( - 2 a _{0} + 1 ) ^{2}}

\int \frac{1}{2 + 4 x ^{2}} d x

n = 0 \sum \infty sin (π \cdot n)

\frac{d}{d x} (6 x x)

x \to - \infty lim (3^{x} x)

\frac{d}{d t} (2 sin (t))

What is the tangent of x3+y3-6xy=0 ?
The tangent of x3+y3-6xy=0 is y=-1/((1-2a_{0))^2}x+(2a_{0}^2)/((-2a_{0)+1)^2}