Question

If \( x = \frac{1 - t^2}{1 + t^2} \) and \( y = \frac{2t}{1 + t^2} \), then \( \frac{dy}{dx} \) is equal to:

• A. \( -\frac{y}{x} \)
• B. \( \frac{y}{x} \)
• C. \( -\frac{x}{y} \)
• D. \( \frac{x}{y} \)

Hint:

When differentiating parametric equations, use the chain rule carefully and substitute back to simplify the result.

Answer 1

The Correct Option is C

We are given the parametric equations: \[ x = \frac{1 - t^2}{1 + t^2}, \quad y = \frac{2t}{1 + t^2}. \] Step 1: Apply trigonometric substitution. Let \( t = \tan\theta \), then the expressions for \( x \) and \( y \) become: \[ x = \cos 2\theta, \quad y = \sin 2\theta. \] Step 2: Differentiate with respect to \( \theta \). The derivatives are: \[ \frac{dx}{d\theta} = -2\sin 2\theta, \quad \frac{dy}{d\theta} = 2\cos 2\theta. \] Now, compute \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{2\cos 2\theta}{-2\sin 2\theta} = -\cot 2\theta. \] Step 3: Simplify the expression. Using the trigonometric identities, we find: \[ \frac{dy}{dx} = -\frac{x}{y}. \] Thus, the correct answer is \( \boxed{(3)} \).