What is the tangent of x^2+y^2=100,(-6,8) ?

The tangent of x^2+y^2=100,(-6,8) is y= 3/4 x+25/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of x^2+y^2=100,(-6,8)

tangent of

x^{2} + y^{2} = 100, (- 6, 8)

y = \frac{3}{4} x + \frac{25}{2}

Solution steps

Find the slope of

x^{2} + y^{2} = 100 : \frac{d y}{d x} = - \frac{x}{y}

Compute the slope of

x^{2} + y^{2} = 100

(- 6, 8) : m = \frac{3}{4}

Find the line with slope m=

\frac{3}{4}

and passing through

(- 6, 8) : y = \frac{3}{4} x + \frac{25}{2}

y = \frac{3}{4} x + \frac{25}{2}

\int - x cos (\frac{n x π}{l}) d x

\frac{d}{d x} (4 x^{3} - 3)

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

x^{2} y^{^{''}} + x y^{^{'}} + y = 0

\int 5 sec^{3} (x) d x

What is the tangent of x^2+y^2=100,(-6,8) ?
The tangent of x^2+y^2=100,(-6,8) is y= 3/4 x+25/2