If `x=(1-t^2)/(1+t^2)` and `y=(2at)/(1+t^2)`, then `(dy)/(dx)=`

Options

A. `(a(1-t^2))/(2t)`
B. `(a(t^2-1))/(2t)`
C. `(a(t^2+1))/(2t)`
D. `(a(t^2-1))/(t)`

Explanation

To find \(\frac{dy}{dx}\) given the parametric equations \(x = \frac{1 - t^2}{1 + t^2}\) and \(y = \frac{2at}{1 + t^2}\), we will follow these steps:

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

Given:
\[ x = \frac{1 - t^2}{1 + t^2} \]

Using the quotient rule for differentiation:
\[ \frac{dx}{dt} = \frac{(1 + t^2)(0 - 2t) - (1 - t^2)(2t)}{(1 + t^2)^2} \]

Calculating the numerator:
\[ = -2t(1 + t^2) - 2t(1 - t^2) = -2t - 2t^3 - 2t + 2t^3 = -4t \]

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

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

Given:
\[ y = \frac{2at}{1 + t^2} \]

Using the quotient rule for differentiation:
\[ \frac{dy}{dt} = \frac{(1 + t^2)(2a) - (2at)(2t)}{(1 + t^2)^2} \]

Calculating the numerator:
\[ = 2a(1 + t^2) - 4at^2 = 2a - 2at^2 \]

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

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

Using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]

Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{\frac{2a(1 - t^2)}{(1 + t^2)^2}}{\frac{-4t}{(1 + t^2)^2}} \]

The \((1 + t^2)^2\) terms cancel out:
\[ \frac{dy}{dx} = \frac{2a(1 - t^2)}{-4t} = \frac{a(1 - t^2)}{-2t} \]

### Final Answer
\[ \frac{dy}{dx} = \frac{a(1 - t^2)}{-2t} \]

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