Find `dy/dx if x^2/a^2 + y^2/(b^2-a^2) = 1`

Answer

To find \(\frac{dy}{dx}\) for the equation

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2 - a^2} = 1, \]

we will use implicit differentiation. Here are the steps:

### Step 1: Differentiate both sides of the equation with respect to \(x\).

Starting with the original equation:

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2 - a^2} = 1, \]

we differentiate each term:

- The derivative of \(\frac{x^2}{a^2}\) with respect to \(x\) is \(\frac{2x}{a^2}\).
- The derivative of \(\frac{y^2}{b^2 - a^2}\) with respect to \(x\) is \(\frac{2y}{b^2 - a^2} \cdot \frac{dy}{dx}\) (using the chain rule).

So, differentiating both sides gives us:

\[ \frac{2x}{a^2} + \frac{2y}{b^2 - a^2} \cdot \frac{dy}{dx} = 0. \]

### Step 2: Rearrange the equation to isolate \(\frac{dy}{dx}\).

We can rearrange the equation to solve for \(\frac{dy}{dx}\):

\[ \frac{2y}{b^2 - a^2} \cdot \frac{dy}{dx} = -\frac{2x}{a^2}. \]

### Step 3: Solve for \(\frac{dy}{dx}\).

Now, divide both sides by \(\frac{2y}{b^2 - a^2}\):

\[ \frac{dy}{dx} = -\frac{2x}{a^2} \cdot \frac{b^2 - a^2}{2y}. \]

The \(2\) cancels out:

\[ \frac{dy}{dx} = -\frac{x(b^2 - a^2)}{a^2y}. \]

### Final Result:

Thus, the derivative \(\frac{dy}{dx}\) is:

\[ \frac{dy}{dx} = -\frac{x(b^2 - a^2)}{a^2y}. \]

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