If \(x = a\left(t - \displaystyle\frac{1}{t}\right),\ y = b\left(t + \displaystyle\frac{1}{t}\right)\), then \(\displaystyle\frac{dy}{dx} =\)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{b^2x}{a^2y}\)
Using the parametric derivative \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\):
\( \displaystyle \frac{dx}{dt} = a\left(1 + \frac{1}{t^2}\right) = \frac{a(t^2 + 1)}{t^2} \)
\( \displaystyle \frac{dy}{dt} = b\left(1 - \frac{1}{t^2}\right) = \frac{b(t^2 - 1)}{t^2} \)
\( \displaystyle \frac{dy}{dx} = \frac{b(t^2 - 1)}{a(t^2 + 1)} \)
To write this in x and y, note that
\( \displaystyle \frac{x}{a} = \frac{t^2 - 1}{t},\quad \frac{y}{b} = \frac{t^2 + 1}{t} \)
Dividing these two:
\( \displaystyle \frac{t^2 - 1}{t^2 + 1} = \frac{x/a}{y/b} = \frac{bx}{ay} \)
So
\( \displaystyle \frac{dy}{dx} = \frac{b}{a} \cdot \frac{bx}{ay} = \frac{b^2x}{a^2y} \)