The slope of the normal at any point (x, y), | vedclass

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(D) The slope of the normal is given by $-\frac{dx}{dy} = \frac{x^2}{xy - x^2y^2 - 1}$.Rearranging the terms,we get $x^2 dy = -xy dx + x^2y^2 dx + dx$

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$\frac{1+	an(1)}{1-	an(1)}$

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(D) The slope of the normal is given by $-\frac{dx}{dy} = \frac{x^2}{xy - x^2y^2 - 1}$.Rearranging the terms,we get $x^2 dy = -xy dx + x^2y^2 dx + dx$.$x^2 dy + xy dx = (x^2y^2 + 1) dx$.$x(x dy + y dx) = (x^2y^2 + 1) dx$.$x d(xy) = (1 + (xy)^2) dx$.Dividing both sides by $x(1 + (xy)^2)$,we get $\frac{d(xy)}{1 + (xy)^2} = \frac{dx}{x}$.Integrating both sides,we obtain $	an^{-1}(xy) = \ln(x) + C$.Since the curve passes through $(1, 1)$,we have $	an^{-1}(1) = \ln(1) + C$,which gives $C = \frac{\pi}{4}$.Thus,$	an^{-1}(xy) = \ln(x) + \frac{\pi}{4}$.Taking the tangent of both sides,$xy = 	an\left(\ln(x) + \frac{\pi}{4}ight) = \frac{	an(\ln x) + 	an(\pi/4)}{1 - 	an(\ln x)	an(\pi/4)} = \frac{1 + 	an(\ln x)}{1 - 	an(\ln x)}$.For $x = e$,$e \cdot y(e) = \frac{1 + 	an(\ln e)}{1 - 	an(\ln e)} = \frac{1 + 	an(1)}{1 - 	an(1)}$.

The slope of the normal at any point $(x, y), x > 0, y > 0$ on the curve $y=y(x)$ is given by $\frac{x^{2}}{x y-x^{2} y^{2}-1}$. If the curve passes through the point $(1, 1)$,then $e \cdot y(e)$ is equal to

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