Q.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 $\frac{2x}{y^2}$.

Concept: Initial Value Problem — we integrate the derivative and then use the given point to find the constant.

Step 1: The slope is $\frac{dy}{dx} = \frac{2x}{y^2}$. Separate variables:

$y^2 \, dy = 2x \, dx$.

Step 2: Integrate both sides:

$\int y^2 \, dy = \int 2x \, dx \quad\Rightarrow\quad \frac{y^3}{3} = x^2 + C$.

Step 3: Use the point $(-2, 3)$: …

The problem gives us the slope of the tangent at any point $(x, y)$ on the curve. That slope is just the derivative $\frac{dy}{dx}$. So we have a first-order differential equation:

and we also know that the curve passes through $(-2, 3)$. This is an Initial Value Problem (IVP): a differential equation plus a specific point that pins down the one particular curve among infinitely many.

The key idea: because the equation is separable — we can move all $y$ terms to one side and all $x$ terms to the other — we can integrate each side separately. That gives us a relationship between $x$ and $y$, and then we use the given point to find the constant of integration.

Let's work through it.

1. Separate the variables. Multiply both sides by $y^2$ and by $dx$:

This is valid as long as $y \neq 0$, which is fine since our point has $y=3$.

2. Integrate both sides.

The left side integrates to $\frac{y^3}{3}$, the right side to $x^2$. Don't forget the constant of integration — put it on one side only:

3. Simplify the equation. Multiply through by 3:

Since $3C$ is just another constant, we can rename it $C$ (or $k$). So:

4. Use the given point to find $C$. The curve passes through $(-2, 3)$. Substitute $x = -2$, $y = 3$: …

Method: From a slope condition to a particular curve

Use this whenever a question describes the slope of the tangent at a general point $(x,y)$ and gives one point the curve passes through. "Slope of the tangent" is simply $\dfrac{dy}{dx}$, so the sentence is a differential equation in disguise.

Steps

Step 1: Translate the slope statement into $\dfrac{dy}{dx} = \text{(given expression)}$.

Write the described slope as the derivative. This is the modelling step that turns words into an equation.

Step 2: Separate and integrate for the general solution. …

Common Mistakes

Mistake 1: Sign error when squaring the negative coordinate.

Why it's wrong: at $(-2,3)$, $(-2)^2 = 4$, not $-4$; a sign slip gives the wrong constant. Correct approach: square carefully to get $27 = 12 + C$, so $C = 15$.

Mistake 2: Forgetting the constant of integration.

Why it's wrong: without $C$ you get $y^3 = 3x^2$, which does not pass through $(-2,3)$. Correct approach: integrate with $+C$ and fix it using the given point. …