Question

If $x=a(t-\frac{1}{t}),y=a(t+\frac{1}{t})$ where $t$ be the parameter then $\frac{dy}{dx}=?$

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

Answer 1

Answer: (C)

$x=a\left(t-\frac{1}{t}\right),y=a\left(t+\frac{1}{t}\right)$
$y^2-x^2=a^2\left[{\left(t+\frac{1}{t}\right)}^2-{\left(t-\frac{1}{t}\right)}^2\right]$
$y^2-x^2=4a^2$
Differentiate w.r.t.$x$
$2y\frac{dy}{dx}-2x=0$
$\therefore \frac{dy}{dx}=\frac{x}{y}$