What is the tangent of x^2+xy-y^2=31,(5,2) ?

The tangent of x^2+xy-y^2=31,(5,2) is y=-12x+62

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tangent of x^2+xy-y^2=31,(5,2)

tangent of

x^{2} + x y - y^{2} = 31, (5, 2)

y = - 12 x + 62

Solution steps

Find the slope of

x^{2} + x y - y^{2} = 31 : \frac{d y}{d x} = \frac{- 2 x - y}{x - 2 y}

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 = - 12

Find the line with slope m=

- 12

and passing through

(5, 2) : y = - 12 x + 62

y = - 12 x + 62

\frac{\partial}{\partial x} (\frac{5 ( 1 - x )}{e ^{x}})

\frac{d}{d x} (1 + 3 cos (2 x))

(ln (x^{2} - 1))^{^{'}}

y^{^{''}} - y^{^{'}} + 30 y = 0

3 sin (2 t) - 3 cos (2 t)

What is the tangent of x^2+xy-y^2=31,(5,2) ?
The tangent of x^2+xy-y^2=31,(5,2) is y=-12x+62