What is the tangent of x^2-y^2=16,(5,3) ?

The tangent of x^2-y^2=16,(5,3) is y= 5/3 x-16/3

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tangent of x^2-y^2=16,(5,3)

tangent of

x^{2} - y^{2} = 16, (5, 3)

y = \frac{5}{3} x - \frac{16}{3}

Solution steps

Find the slope of

x^{2} - y^{2} = 16 : \frac{d y}{d x} = \frac{x}{y}

Compute the slope of

x^{2} - y^{2} = 16

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

Find the line with slope m=

\frac{5}{3}

and passing through

(5, 3) : y = \frac{5}{3} x - \frac{16}{3}

y = \frac{5}{3} x - \frac{16}{3}

\int \frac{2 x ^{2} + x + 4}{x ^{3} + 4 x ^{2}} d x

\int \frac{6}{sin ^{2} ( x )} d x

\frac{4 x}{x - 3}, at x = 5

f (x) = \frac{x ^{3}}{3} + x^{2} + 4 x - 10

y = 10 x - x^{2}, (5, 26)

What is the tangent of x^2-y^2=16,(5,3) ?
The tangent of x^2-y^2=16,(5,3) is y= 5/3 x-16/3