Question

If  \[x = a\left( t + \frac{1}{t} \right) \text{ and y } = a\left( t - \frac{1}{t} \right)\] ,prove that  \[\frac{dy}{dx} = \frac{x}{y}\]?

 

Answer 1

\[\text{ We have, x } = a\left( t + \frac{1}{t} \right) \text{ and y } = a\left( t - \frac{1}{t} \right)\]

\[\Rightarrow \frac{dx}{dt} = a\frac{d}{dt}\left( t + \frac{1}{t} \right) \text{ and } \frac{dy}{dt} = a\frac{d}{dt}\left( t - \frac{1}{t} \right)\]
\[ \Rightarrow \frac{dx}{dt} = a\left( 1 - \frac{1}{t^2} \right) \text{ and } \frac{dy}{dt} = a\left( 1 + \frac{1}{t^2} \right) \]
\[ \Rightarrow \frac{dx}{dt} = a\left( \frac{t^2 - 1}{t^2} \right) \text{ and } \frac{dy}{dt} = a\left( \frac{t^2 + 1}{t^2} \right) \]
\[ \therefore \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{a\left( t^2 + 1 \right)}{t^2} \times \frac{t^2}{a\left( t^2 - 1 \right)}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{a\left( t^2 + 1 \right)}{t} \times \frac{t}{a\left( t^2 - 1 \right)}\]
\[ \Rightarrow \frac{dy}{dx} = a\left( t + \frac{1}{t} \right) \times \frac{1}{a\left( t - \frac{1}{t} \right)}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{x}{y} \]