Find `dy/dx`, if x and y are connected parametrically by the equations, given below without eliminating the parameter:
`x=(a(1+t^(2)))/(1-t^(2)),y=(2t)/(1-t^(2))`

Answer

To find \( \frac{dy}{dx} \) when \( x \) and \( y \) are connected parametrically by the equations:

\[ x = \frac{a(1+t^2)}{1-t^2}, \quad y = \frac{2t}{1-t^2} \]

we will use the chain rule without eliminating the parameter \( t \).

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

We start by differentiating \( x \) with respect to \( t \):

\[ x = \frac{a(1+t^2)}{1-t^2} \]

Using the quotient rule, where \( u = a(1+t^2) \) and \( v = 1-t^2 \):

\[ \frac{dx}{dt} = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2} \]

Calculating \( \frac{du}{dt} \) and \( \frac{dv}{dt} \):

\[ \frac{du}{dt} = a(2t), \quad \frac{dv}{dt} = -2t \]

Now substituting these into the quotient rule:

\[ \frac{dx}{dt} = \frac{(1-t^2)(2at) - a(1+t^2)(-2t)}{(1-t^2)^2} \]

Simplifying this expression:

\[ \frac{dx}{dt} = \frac{2at(1-t^2) + 2at(1+t^2)}{(1-t^2)^2} = \frac{2at(1-t^2 + 1+t^2)}{(1-t^2)^2} = \frac{2at(2)}{(1-t^2)^2} = \frac{4at}{(1-t^2)^2} \]

### Step 2: Differentiate \( y \) with respect to \( t \)

Next, we differentiate \( y \) with respect to \( t \):

\[ y = \frac{2t}{1-t^2} \]

Using the quotient rule again:

\[ \frac{dy}{dt} = \frac{(1-t^2)(2) - 2t(-2t)}{(1-t^2)^2} \]

Simplifying this expression:

\[ \frac{dy}{dt} = \frac{2(1-t^2) + 4t^2}{(1-t^2)^2} = \frac{2 + 2t^2}{(1-t^2)^2} = \frac{2(1+t^2)}{(1-t^2)^2} \]

### Step 3: Find \( \frac{dy}{dx} \)

Now, we can find \( \frac{dy}{dx} \) using the chain rule:

\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]

Substituting the derivatives we found:

\[ \frac{dy}{dx} = \frac{\frac{2(1+t^2)}{(1-t^2)^2}}{\frac{4at}{(1-t^2)^2}} \]

The \( (1-t^2)^2 \) terms cancel out:

\[ \frac{dy}{dx} = \frac{2(1+t^2)}{4at} = \frac{1+t^2}{2at} \]

### Final Result

Thus, the final result is:

\[ \frac{dy}{dx} = \frac{1+t^2}{2at} \]

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