# What is the slope of the tangent line of  (x^2/y-y^2)/(1+x)^2 =C , where C is an arbitrary constant, at (6,5)?

Apr 14, 2018

Slope is $\frac{3495}{648}$ and tangent is $y - 5 = \frac{3495}{648} \left(x - 6\right)$

#### Explanation:

As we are seeking tangent at point $\left(6 , 5\right)$, the curve $\frac{{x}^{2} / y - {y}^{2}}{1 + x} ^ 2 = C$ passes through it and hence

$C = \frac{{x}^{2} / y - {y}^{2}}{1 + x} ^ 2 = \frac{{6}^{2} / 5 - {5}^{2}}{1 + 5} ^ 2 = - \frac{89}{5 \cdot 36} = - \frac{89}{180}$

i.e. curve is $\frac{{x}^{2} / y - {y}^{2}}{1 + x} ^ 2 = - \frac{89}{180}$

or ${x}^{2} / y - {y}^{2} = - \frac{89}{180} {\left(1 + x\right)}^{2}$

Slope of tangent is given by value of $\frac{\mathrm{dy}}{\mathrm{dx}}$ at this point

and differentiating we get

$2 \frac{x}{y} - {x}^{2} / {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{89}{180} \left(2 x + 2\right)$

or at $\left(6 , 5\right)$ we have $\frac{12}{5} - \frac{36}{25} \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{89}{180} \cdot 14$

or $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{25}{36} \left(\frac{12}{5} + \frac{483}{90}\right) = \frac{25}{36} \cdot \frac{699}{90} = \frac{3495}{648}$

and tangent is $y - 5 = \frac{3495}{648} \left(x - 6\right)$