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Question

# Find the equation of the curve that passes through the point (0, a) and is such that at any point (x, y) on it, the product of its slope and the ordinate is equal to the abscissa.

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Solution

## According to the question, $y\frac{dy}{dx}=x\phantom{\rule{0ex}{0ex}}⇒ydy=xdx\phantom{\rule{0ex}{0ex}}\mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}x,\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}\int ydy=\int xdx\phantom{\rule{0ex}{0ex}}⇒\frac{{y}^{2}}{2}=\frac{{x}^{2}}{2}+C\phantom{\rule{0ex}{0ex}}\mathrm{Since}\mathrm{the}\mathrm{curve}\mathrm{passes}\mathrm{through}\left(0,a\right),\mathrm{it}\mathrm{satisfies}\mathrm{the}\mathrm{above}\mathrm{equation}.\phantom{\rule{0ex}{0ex}}\therefore \frac{{a}^{2}}{2}=\frac{0}{2}+C\phantom{\rule{0ex}{0ex}}⇒C=\frac{{a}^{2}}{2}\phantom{\rule{0ex}{0ex}}\mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}C,\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}\frac{{y}^{2}}{2}=\frac{{x}^{2}}{2}+\frac{{a}^{2}}{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-{y}^{2}=-{a}^{2}$

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