What is the normal of y= x/(sqrt(4+x^2)),(0,0) ?

The normal of y= x/(sqrt(4+x^2)),(0,0) is y=-2x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

normal of y= x/(sqrt(4+x^2)),(0,0)

normal of

y = \frac{x}{4 + x ^{2}}, (0, 0)

y = - 2 x

Solution steps

Find the slope of

y = \frac{x}{4 + x ^{2}} : \frac{d y}{d x} = \frac{4}{( 4 + x ^{2} ) 4 + 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}{2}

Compute the slope of the perpendicular line:

m_{p} = - 2

Find the line with slope m=

- 2

and passing through

(0, 0) : y = - 2 x

y = - 2 x

What is the normal of y= x/(sqrt(4+x^2)),(0,0) ?
The normal of y= x/(sqrt(4+x^2)),(0,0) is y=-2x