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) → \( \displaystyle\frac{b^2x}{a^2y} \)
By parametric differentiation, \( \frac{dy}{dx}=\frac{dy/dt}{dx/dt} \).
\( \displaystyle \frac{dx}{dt}=a\left(1+\frac{1}{t^2}\right)=\frac{a(t^2+1)}{t^2} \)
\( \displaystyle \frac{dy}{dt}=b\left(1-\frac{1}{t^2}\right)=\frac{b(t^2-1)}{t^2} \)
\( \displaystyle \frac{dy}{dx}=\frac{b(t^2-1)}{a(t^2+1)} \)
To write this in x and y, note that
\( \displaystyle \frac{x}{a}=\frac{t^2-1}{t},\qquad \frac{y}{b}=\frac{t^2+1}{t} \)
\( \displaystyle \frac{t^2-1}{t^2+1}=\frac{x/a}{y/b}=\frac{bx}{ay} \)
Hence
\( \displaystyle \frac{dy}{dx}=\frac{b}{a}\cdot\frac{bx}{ay} \)
\( \displaystyle =\frac{b^2x}{a^2y} \)
Shortcut: Eliminate t first:
\( \displaystyle \frac{y^2}{b^2}-\frac{x^2}{a^2}=\left(t+\frac{1}{t}\right)^2-\left(t-\frac{1}{t}\right)^2=4 \)
Differentiating, \( \frac{2y}{b^2}\frac{dy}{dx}-\frac{2x}{a^2}=0 \), so \( \frac{dy}{dx}=\frac{b^2x}{a^2y} \).
Why the other options are incorrect
Option (1) \( \displaystyle\frac{x}{y} \) — Incorrect. \( \frac{x}{y} \) is the answer only when a = b; the factor \( \frac{b^2}{a^2} \) is lost.
Option (3) \( \displaystyle\frac{bx}{ay} \) — Incorrect. \( \frac{bx}{ay} \) is only the ratio \( \frac{t^2-1}{t^2+1} \); it still has to be multiplied by \( \frac{b}{a} \).
Option (4) \( \displaystyle\frac{a^2y}{b^2x} \) — Incorrect. \( \frac{a^2y}{b^2x} \) is \( \frac{dx}{dy} \), the reciprocal of the answer.