Question

If $x = a\left(t - \frac{1}{t}\right)$ and $y = b\left(t + \frac{1}{t}\right)$, then $\frac{dy}{dx} =$

• A. $\frac{a^2 x}{b^2 y}$
• B. $\frac{a^2 y}{b^2 x}$
• C. $-\frac{b^2 x}{a^2 y}$
• D. $\frac{b^2 x}{a^2 y}$

Hint:

Eliminating parameters using identities is an incredible shortcut for problems containing $(t + \frac{1}{t})$ and $(t - \frac{1}{t})$. Recognizing that it models a hyperbola of the form $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 4$ allows you to differentiate implicitly, completely skipping complex quotient rule work with parameter fractions!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given a pair of parametric equations where $x$ and $y$ are expressed in terms of an underlying parameter $t$. We need to compute the derivative $\frac{dy}{dx}$ and express the final result in terms of the variables $x$ and $y$.

Step 2: Key Formula or Approach:
1. Use parametric differentiation rules to compute individual derivatives with respect to $t$: $$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$ 2. Alternatively, eliminate the parameter $t$ by squaring and combining the equations to exploit the algebraic identity: $$\left(t + \frac{1}{t}\right)^2 - \left(t - \frac{1}{t}\right)^2 = 4$$ 3. Perform implicit differentiation on the resulting equation.

Step 3: Detailed Explanation:
Let's isolate the parameter terms from the given equations: $$\frac{x}{a} = t - \frac{1}{t} \quad \text{--- (1)}$$ $$\frac{y}{b} = t + \frac{1}{t} \quad \text{--- (2)}$$ Square both expressions independently: $$\frac{x^2}{a^2} = \left(t - \frac{1}{t}\right)^2 = t^2 - 2 + \frac{1}{t^2}$$ $$\frac{y^2}{b^2} = \left(t + \frac{1}{t}\right)^2 = t^2 + 2 + \frac{1}{t^2}$$ Subtract the squared $x$ expression from the squared $y$ expression: $$\frac{y^2}{b^2} - \frac{x^2}{a^2} = \left(t^2 + 2 + \frac{1}{t^2}\right) - \left(t^2 - 2 + \frac{1}{t^2}\right)$$ $$\frac{y^2}{b^2} - \frac{x^2}{a^2} = 4$$ Now, differentiate this hyperbola equation implicitly with respect to $x$: $$\frac{d}{dx}\left(\frac{y^2}{b^2}\right) - \frac{d}{dx}\left(\frac{x^2}{a^2}\right) = \frac{d}{dx}(4)$$ $$\frac{2y}{b^2} \cdot \frac{dy}{dx} - \frac{2x}{a^2} = 0$$ Isolate the derivative term $\frac{dy}{dx}$: $$\frac{2y}{b^2} \cdot \frac{dy}{dx} = \frac{2x}{a^2}$$ $$\frac{dy}{dx} = \frac{2x}{a^2} \times \frac{b^2}{2y} = \frac{b^2 x}{a^2 y}$$

Step 4: Final Answer:
The derivative expression matches option (D).