What is the tangent of x^2+xy-y^2=-1,(3,5) ?

The tangent of x^2+xy-y^2=-1,(3,5) is y= 11/7 x+2/7

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tangent of x^2+xy-y^2=-1,(3,5)

tangent of

x^{2} + x y - y^{2} = - 1, (3, 5)

y = \frac{11}{7} x + \frac{2}{7}

Solution steps

Find the slope of

x^{2} + x y - y^{2} = - 1 : \frac{d y}{d x} = \frac{- 2 x - y}{x - 2 y}

Compute the slope of

x^{2} + x y - y^{2} = - 1

(3, 5) : m = \frac{11}{7}

Find the line with slope m=

\frac{11}{7}

and passing through

(3, 5) : y = \frac{11}{7} x + \frac{2}{7}

y = \frac{11}{7} x + \frac{2}{7}

\frac{4 x ^{2} + 4 x + 8}{x}

\int e^{12 x} (12) d x

f (x) = 3 x^{2} cos (x) + 8 x

x \to 0 + lim ((sin (x))^{x})

\int cot^{3} (3 x) d x

What is the tangent of x^2+xy-y^2=-1,(3,5) ?
The tangent of x^2+xy-y^2=-1,(3,5) is y= 11/7 x+2/7