Question

If $x = \frac{1 - t^2}{1 + t^2}$ and $y = \frac{2at}{1 + t^2}$, then $\frac{dy}{dx} =$

• A. $\frac{a(t^2 + 1)}{2t}$
• B. $\frac{a(t^2 - 1)}{t}$
• C. $\frac{a(1 - t^2)}{2t}$
• D. $\frac{a(t^2 - 1)}{2t}$

Hint:

Use trigonometric substitution to bypass the quotient rule entirely! Let $t = \tan\theta$.
Then $x = \frac{1 - \tan^2\theta}{1 + \tan^2\theta} = \cos(2\theta)$ and $y = \frac{2a\tan\theta}{1 + \tan^2\theta} = a\sin(2\theta)$.
Differentiating with respect to $\theta$: $\frac{dx}{d\theta} = -2\sin(2\theta)$ and $\frac{dy}{d\theta} = 2a\cos(2\theta)$.
Thus, $\frac{dy}{dx} = \frac{2a\cos(2\theta)}{-2\sin(2\theta)} = -a\cot(2\theta) = -a \left( \frac{1 - \tan^2\theta}{2\tan\theta} \right) = \frac{a(\tan^2\theta - 1)}{2\tan\theta} = \frac{a(t^2 - 1)}{2t}$.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given two parametric equations where variables $x$ and $y$ are explicitly defined in terms of a real parameter $t$. We need to evaluate the first derivative $\frac{dy}{dx}$ in terms of $t$.

Step 2: Key Formula or Approach:
According to the parametric rule of differentiation: $$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$ We can differentiate both expressions individually with respect to $t$ using the Quotient Rule: $$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$$

Step 3: Detailed Explanation:
Let's first compute $\frac{dx}{dt}$ using the quotient rule on $x = \frac{1 - t^2}{1 + t^2}$: $$\frac{dx}{dt} = \frac{(-2t)(1 + t^2) - (1 - t^2)(2t)}{(1 + t^2)^2}$$ $$\frac{dx}{dt} = \frac{-2t - 2t^3 - 2t + 2t^3}{(1 + t^2)^2} = \frac{-4t}{(1 + t^2)^2}$$ Now, let's compute $\frac{dy}{dt}$ using the quotient rule on $y = \frac{2at}{1 + t^2}$: $$\frac{dy}{dt} = \frac{(2a)(1 + t^2) - (2at)(2t)}{(1 + t^2)^2}$$ $$\frac{dy}{dt} = \frac{2a + 2at^2 - 4at^2}{(1 + t^2)^2} = \frac{2a(1 - t^2)}{(1 + t^2)^2}$$ Now combine both derivatives to find $\frac{dy}{dx}$: $$\frac{dy}{dx} = \frac{\frac{2a(1 - t^2)}{(1 + t^2)^2}}{\frac{-4t}{(1 + t^2)^2}}$$ The common denominators cancel out cleanly: $$\frac{dy}{dx} = \frac{2a(1 - t^2)}{-4t} = \frac{a(1 - t^2)}{-2t}$$ Absorb the negative sign from the denominator into the numerator to flip the terms: $$\frac{dy}{dx} = \frac{a(t^2 - 1)}{2t}$$

Step 4: Final Answer:
The derivative $\frac{dy}{dx}$ simplifies to $\frac{a(t^2 - 1)}{2t}$, which matches option (D).