Question

Find the equation of the curve which passes through the point (3, −4) and has the slope \[\frac{2y}{x}\]  at any point (x, y) on it.

Answer 1

According to the question,
\[\frac{dy}{dx} = \frac{2y}{x}\]
\[\Rightarrow \frac{1}{2y}dy = \frac{1}{x}dx\]
Integrating both sides with respect to x, we get
\[\int\frac{1}{2y}dy = \int\frac{1}{x}dx\]
\[ \Rightarrow \frac{1}{2}\log \left| y \right| = \log \left| x \right| + C\]
\[\text{ Since the curve passes through }\left( 3, - 4 \right),\text{ it satisfies the above equation . }\]
\[ \therefore \frac{1}{2}\log \left| - 4 \right| = \log \left| 3 \right| + C\]
\[ \Rightarrow \log \left| 2 \right| - \log \left| 3 \right| = C\]
\[ \Rightarrow C = \log \left| \frac{2}{3} \right|\]
Putting the value of C, we get
\[\log \left| y \right| = 2\log \left| x \right| + 2\log \left| \frac{2}{3} \right|\]
\[ \Rightarrow \log \left| y \right| = \log \left| \frac{4}{9} x^2 \right|\]
\[ \Rightarrow y = \pm \frac{4}{9} x^2 \]
\[ \Rightarrow 9y - 4 x^2 = 0\text{ or }9y + 4 x^2 = 0\]
\[\text{ The given point does not satisfy the equation }9y - 4 x^2 = 0 . \]
\[ \therefore 9y + 4 x^2 = 0\]