What is the tangent of x^2+xy-y^2=19,(4,1) ?

The tangent of x^2+xy-y^2=19,(4,1) is y=-9/2 x+19

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tangent of x^2+xy-y^2=19,(4,1)

tangent of

x^{2} + x y - y^{2} = 19, (4, 1)

y = - \frac{9}{2} x + 19

Solution steps

Find the slope of

x^{2} + x y - y^{2} = 19 : \frac{d y}{d x} = \frac{- 2 x - y}{x - 2 y}

Compute the slope of

x^{2} + x y - y^{2} = 19

(4, 1) : m = - \frac{9}{2}

Find the line with slope m=

- \frac{9}{2}

and passing through

(4, 1) : y = - \frac{9}{2} x + 19

y = - \frac{9}{2} x + 19

x \to 0 + lim (\frac{cos ( x ) - 1}{x})

\int_{- 3}^{3} 1 + (x)^{2} d x

x^{2} - 4 x + 8

\frac{s + 1}{s ^{2} + 3 s}

x \cdot \frac{d y}{d x} - y = 3 x^{6} y, y (1) = 2

What is the tangent of x^2+xy-y^2=19,(4,1) ?
The tangent of x^2+xy-y^2=19,(4,1) is y=-9/2 x+19