​$$\text { If } x=a\left(t+\frac{1}{t}\right) \text { and } y=a\left(t-\frac{1}{t}\right) \text {, then } \frac{d x}{d y} \text { is: }$$​​

Correct option is A

​$$\textbf{Step 1: Compute } \frac{dx}{dt} \text{ and } \frac{dy}{dt} \\[6pt]\frac{dx}{dt} = a\left(1 - \frac{1}{t^2} \right) = a\left(\frac{t^2 - 1}{t^2} \right) \\[6pt]\frac{dy}{dt} = a\left(1 + \frac{1}{t^2} \right) = a\left(\frac{t^2 + 1}{t^2} \right) \\[12pt]\textbf{Step 2: Find } \frac{dy}{dx} \text{ using the chain rule} \\[6pt]\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a\left(\frac{t^2 + 1}{t^2}\right)}{a\left(\frac{t^2 - 1}{t^2}\right)} = \frac{t^2 + 1}{t^2 - 1} \\[12pt]\textbf{Step 3: Compute } \frac{dx}{dy} \text{ (the reciprocal of } \frac{dy}{dx} \text{)} \\[6pt]\frac{dx}{dy} = \frac{t^2 - 1}{t^2 + 1} \\[12pt]$$​​