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

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

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-y^2=1,(2,1)

tangent of

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

y = \frac{3}{4} x - \frac{1}{2}

Solution steps

Find the slope of

x^{2} - x y - y^{2} = 1 : \frac{d y}{d x} = - \frac{- 2 x + y}{x + 2 y}

Compute the slope of

x^{2} - x y - y^{2} = 1

(2, 1) : m = \frac{3}{4}

Find the line with slope m=

\frac{3}{4}

and passing through

(2, 1) : y = \frac{3}{4} x - \frac{1}{2}

y = \frac{3}{4} x - \frac{1}{2}

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

\frac{d y}{d x} = (\frac{x y + 5 x - y - 5}{x y - 2 x + 6 y - 12})

x y^{3} d x + e^{x^{2}} d y = 0

(5 x^{2} + 2 x^{- 1} y + 2 x^{- 1} y^{2}) d x + (1 + 2 y) d y = 0

e^{θ} sin (5 θ)

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