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Question
Find the equation of a curve passing through the point (−2, 3), given that the slope of the tangent to the curve at any point (x, y) is `(2x)/y^2.`
Solution
We have,
\[\frac{dy}{dx} = \frac{2x}{y^2}\]
\[ \Rightarrow y^2 dy = 2x dx\]
Integrating both sides, we get
\[\int y^2 dy = 2\int x dx\]
\[ \Rightarrow \frac{y^3}{3} = x^2 + C . . . . . \left( 1 \right)\]
Now the given curve passes theough (- 2, 3)
Therefore, when x = - 2, y = 3
Substituting x = - 2 and y = 3 in (1) we get
\[\frac{3^3}{3} = \left( - 2 \right)^2 + C\]
\[ \Rightarrow 9 = 4 + C\]
\[ \Rightarrow C = 5\]
Putting the value of `C` in (1), we get
\[\frac{y^3}{3} = x^2 + 5\]
\[ \Rightarrow y^3 = 3 x^2 + 15\]
\[ \Rightarrow y = \left( 3 x^2 + 15 \right)^\frac{1}{3}\]
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