Show that the family of curves for which the slope of the tangent at any point $(x, y)$ on it is $\frac{x^{2}+y^{2}}{2 x y}$ is given by $x^{2}-y^{2}=c x$.

(A) We know that the slope of the tangent at any point on a curve is $\frac{dy}{dx}$.
Therefore,$\frac{dy}{dx} = \frac{x^{2}+y^{2}}{2xy}$.
Dividing the numerator and denominator by $x^{2}$,we get $\frac{dy}{dx} = \frac{1+(y/x)^{2}}{2(y/x)}$. This is a homogeneous differential equation.
Let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
Substituting these into the equation: $v + x\frac{dv}{dx} = \frac{1+v^{2}}{2v}$.
$x\frac{dv}{dx} = \frac{1+v^{2}}{2v} - v = \frac{1+v^{2}-2v^{2}}{2v} = \frac{1-v^{2}}{2v}$.
Separating the variables: $\frac{2v}{1-v^{2}} dv = \frac{dx}{x}$.
Integrating both sides: $\int \frac{2v}{1-v^{2}} dv = \int \frac{dx}{x}$.
Let $1-v^{2} = t$,then $-2v dv = dt$,so $\int -\frac{dt}{t} = \ln|x| + C$.
$-\ln|1-v^{2}| = \ln|x| + C$.
$\ln|1-v^{2}|^{-1} = \ln|x| + C \implies \frac{1}{1-v^{2}} = Cx$.
Substituting $v = y/x$: $\frac{1}{1-(y^{2}/x^{2})} = Cx \implies \frac{x^{2}}{x^{2}-y^{2}} = Cx$.
$x^{2} = C x (x^{2}-y^{2}) \implies x = C(x^{2}-y^{2}) \implies x^{2}-y^{2} = \frac{1}{C} x = cx$.