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Question

# The slope of the tangent to the curve at any point is the reciprocal of twice the ordinate at that point. The curve passes through the point (4, 3). Determine its equation.

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Solution

## According to the question, $\frac{dy}{dx}=\frac{1}{2y}\phantom{\rule{0ex}{0ex}}⇒2ydy=dx\phantom{\rule{0ex}{0ex}}\mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}2\int ydy=\int dx\phantom{\rule{0ex}{0ex}}⇒{y}^{2}=x+C\phantom{\rule{0ex}{0ex}}\mathrm{Since}\mathrm{the}\mathrm{curve}\mathrm{passes}\mathrm{throught}\mathrm{the}\mathrm{point}\left(4,3\right),\mathrm{it}\mathrm{satisfies}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{curve}.\phantom{\rule{0ex}{0ex}}9=4+C\phantom{\rule{0ex}{0ex}}⇒C=5\phantom{\rule{0ex}{0ex}}\mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}C\mathrm{in}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{curve},\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}{y}^{2}=x+5$

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