What is the tangent of x^2-3xy=10 ?

The tangent of x^2-3xy=10 is y=-(-a_{0}^2-10)/(3a_{0)^2}x-(20)/(3a_{0)}

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tangent of x^2-3xy=10

tangent of

x^{2} - 3 x y = 10

y = - \frac{- a _{0}^{2} - 10}{3 a _{0}^{2}} x - \frac{20}{3 a _{0}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - \frac{10 - a _{0}^{2}}{3 a _{0}})

Find the slope of

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

Compute the slope of

x^{2} - 3 x y = 10

x = a_{0} : m = - \frac{- a _{0}^{2} - 10}{3 a _{0}^{2}}

Find the line with slope m=

- \frac{- a _{0}^{2} - 10}{3 a _{0}^{2}}

and passing through

(a_{0}, - \frac{10 - a _{0}^{2}}{3 a _{0}}) : y = - \frac{- a _{0}^{2} - 10}{3 a _{0}^{2}} x - \frac{20}{3 a _{0}}

y = - \frac{- a _{0}^{2} - 10}{3 a _{0}^{2}} x - \frac{20}{3 a _{0}}

y^{^{'}} = \frac{sec ( y )}{( x - 3 ) ^{2}}

n = 0 \sum \infty \frac{2}{1}

\frac{\partial}{\partial x} (4 x + y)

\frac{d}{d x} ((\frac{2}{x} - \frac{1}{x ^{2}})^{\frac{1}{2}})

\int 2 tan^{4} (x) sec^{6} (x) d x

What is the tangent of x^2-3xy=10 ?
The tangent of x^2-3xy=10 is y=-(-a_{0}^2-10)/(3a_{0)^2}x-(20)/(3a_{0)}