What is the tangent of (10-x)y^2=x^3,(2,1) ?

The tangent of (10-x)y^2=x^3,(2,1) is y= 13/16 x-5/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

tangent of (10-x)y^2=x^3,(2,1)

tangent of

(10 - x) y^{2} = x^{3}, (2, 1)

y = \frac{13}{16} x - \frac{5}{8}

Solution steps

Find the slope of

(10 - x) y^{2} = x^{3} : \frac{d y}{d x} = \frac{3 x ^{2} + y ^{2}}{2 y ( 10 - x )}

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{13}{16}

Find the line with slope m=

\frac{13}{16}

and passing through

(2, 1) : y = \frac{13}{16} x - \frac{5}{8}

y = \frac{13}{16} x - \frac{5}{8}

What is the tangent of (10-x)y^2=x^3,(2,1) ?
The tangent of (10-x)y^2=x^3,(2,1) is y= 13/16 x-5/8