What is the tangent of 4x^2+xy+4y^2=9,(1,1) ?

The tangent of 4x^2+xy+4y^2=9,(1,1) is y=-x+2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of 4x^2+xy+4y^2=9,(1,1)

tangent of

4 x^{2} + x y + 4 y^{2} = 9, (1, 1)

y = - x + 2

Solution steps

Find the slope of

4 x^{2} + x y + 4 y^{2} = 9 : \frac{d y}{d x} = \frac{- 8 x - y}{x + 8 y}

Compute the slope of

4 x^{2} + x y + 4 y^{2} = 9

(1, 1) : m = - 1

Find the line with slope m=

- 1

and passing through

(1, 1) : y = - x + 2

y = - x + 2

\frac{d}{d x} (3 x cot (4 x - 2))

y = x^{2} - 1, y = 8

\frac{1}{x}, \frac{5}{x}, x, 3 x

\int_{- π}^{π} sin (3 x) d x

x^{2} + y e^{x} + (e^{x} - y) (\frac{d y}{d x}) = 0

What is the tangent of 4x^2+xy+4y^2=9,(1,1) ?
The tangent of 4x^2+xy+4y^2=9,(1,1) is y=-x+2