If `x=a(t+1/t)` and `y=a(t-1/t)` , prove that `(dy)/(dx)=x/y`

Explanation

To prove that \(\frac{dy}{dx} = \frac{x}{y}\) given the equations \(x = a\left(t + \frac{1}{t}\right)\) and \(y = a\left(t - \frac{1}{t}\right)\), we can follow these steps:

### Step 1: Differentiate \(x\) and \(y\) with respect to \(t\)

Given:
\[ x = a\left(t + \frac{1}{t}\right) \]
Differentiate \(x\) with respect to \(t\):
\[ \frac{dx}{dt} = a\left(1 - \frac{1}{t^2}\right) \]

Now for \(y\):
\[ y = a\left(t - \frac{1}{t}\right) \]
Differentiate \(y\) with respect to \(t\):
\[ \frac{dy}{dt} = a\left(1 + \frac{1}{t^2}\right) \]

### Step 2: Use the chain rule to find \(\frac{dy}{dx}\)

Using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{a\left(1 + \frac{1}{t^2}\right)}{a\left(1 - \frac{1}{t^2}\right)} \]
The \(a\) cancels out:
\[ \frac{dy}{dx} = \frac{1 + \frac{1}{t^2}}{1 - \frac{1}{t^2}} \]

### Step 3: Simplify the expression

To simplify, multiply the numerator and denominator by \(t^2\):
\[ \frac{dy}{dx} = \frac{t^2 + 1}{t^2 - 1} \]

### Step 4: Express \(x\) and \(y\) in terms of \(t\)

Recall:
\[ x = a\left(t + \frac{1}{t}\right) \quad \text{and} \quad y = a\left(t - \frac{1}{t}\right) \]
We can express \(t + \frac{1}{t}\) and \(t - \frac{1}{t}\) in terms of \(x\) and \(y\):
\[ t + \frac{1}{t} = \frac{x}{a} \quad \text{and} \quad t - \frac{1}{t} = \frac{y}{a} \]

### Step 5: Find \(x/y\)

Now, we need to find \(\frac{x}{y}\):
\[ \frac{x}{y} = \frac{a\left(t + \frac{1}{t}\right)}{a\left(t - \frac{1}{t}\right)} = \frac{t + \frac{1}{t}}{t - \frac{1}{t}} = \frac{t^2 + 1}{t^2 - 1} \]

### Step 6: Conclude the proof

Since we have:
\[ \frac{dy}{dx} = \frac{t^2 + 1}{t^2 - 1} \quad \text{and} \quad \frac{x}{y} = \frac{t^2 + 1}{t^2 - 1} \]
We conclude that:
\[ \frac{dy}{dx} = \frac{x}{y} \]

Thus, we have proved that \(\frac{dy}{dx} = \frac{x}{y}\).