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.`

Answer 1

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

\[\frac{y^3}{3} = x^2 + 5\]

\[ \Rightarrow y^3 = 3 x^2 + 15\]

\[ \Rightarrow y = \left( 3 x^2 + 15 \right)^\frac{1}{3}\]