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

Options

A. ` (x)/(y)`
B. ` (-x)/( y)`
C. ` (b^(2) x)/( a^(2) y)`
D. ` (-b^(2)x)/( a^(2)y)`

Answer

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

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

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

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

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

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

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

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

Again using the quotient rule:
\[ \frac{dy}{dt} = \frac{(1 + t^2)(2b) - (2bt)(2t)}{(1 + t^2)^2} \]

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

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

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

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

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

The \((1 + t^2)^2\) cancels out:
\[ \frac{dy}{dx} = \frac{2b(1 - t^2)}{-4at} = \frac{-b(1 - t^2)}{2at} \]

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

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