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 can use the formula:

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

### Step 1: Find \(\frac{dx}{dt}\)

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

Using the quotient rule for differentiation, where \(f(t) = 1 - t^2\) and \(g(t) = 1 + t^2\):

\[ \frac{dx}{dt} = \frac{f' \cdot g - f \cdot g'}{g^2} \]

Calculating \(f'\) and \(g'\):
- \(f' = -2t\)
- \(g' = 2t\)

Now substituting into the quotient rule:

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

Expanding the numerator:

\[ = \frac{-2t - 2t^3 - 2t + 2t^3}{(1 + t^2)^2} \]
\[ = \frac{-4t}{(1 + t^2)^2} \]

### Step 2: Find \(\frac{dy}{dt}\)

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

Using the quotient rule again, where \(f(t) = 2at\) and \(g(t) = 1 + t^2\):

Calculating \(f'\) and \(g'\):
- \(f' = 2a\)
- \(g' = 2t\)

Now substituting into the quotient rule:

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

Expanding the numerator:

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

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

Now substituting \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\) into the formula:

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

The \((1 + t^2)^2\) cancels out:

\[ = \frac{2a(1 - t^2)}{-4t} \]
\[ = -\frac{a(1 - t^2)}{2t} \]

Thus, the final result is:

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