Match List-I with List-II, where \(0 \lt t \lt \displaystyle\frac{\pi}{2}\)
| List-I | List-II | ||
|---|---|---|---|
| Parametric equations | \(\displaystyle\frac{dy}{dx}\) | ||
| (A) | \(x = 3\cos t,\ y = 3\sin t\) | (I) | \(\mathrm{cosec}\,t\) |
| (B) | \(x = e^{t},\ y = e^{-t}\) | (II) | \(-\cot t\) |
| (C) | \(x = t + \displaystyle\frac{1}{t},\ y = t - \displaystyle\frac{1}{t}\) | (III) | \(\displaystyle\frac{t^{2} + 1}{t^{2} - 1}\) |
| (D) | \(x = \sec t,\ y = \tan t\) | (IV) | \(-e^{-2t}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(II), (B)-(IV), (C)-(III), (D)-(I)
In parametric form, \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\) wherever \(\frac{dx}{dt} \ne 0\).
(A) \(x = 3\cos t,\ y = 3\sin t\)
This matches (II) \(-\cot t\).
\( \displaystyle \frac{dy}{dx} = \frac{3\cos t}{-3\sin t} = -\cot t \)
(B) \(x = e^{t},\ y = e^{-t}\)
This matches (IV) \(-e^{-2t}\).
\( \displaystyle \frac{dy}{dx} = \frac{-e^{-t}}{e^{t}} = -e^{-2t} \)
(C) \(x = t + \frac{1}{t},\ y = t - \frac{1}{t}\)
This matches (III) \(\frac{t^{2} + 1}{t^{2} - 1}\).
\( \displaystyle \frac{dx}{dt} = 1 - \frac{1}{t^{2}} \)
\( \displaystyle \frac{dy}{dt} = 1 + \frac{1}{t^{2}} \)
Multiplying the top and bottom by \(t^{2}\):
\( \displaystyle \frac{dy}{dx} = \frac{t^{2} + 1}{t^{2} - 1} \)
(D) \(x = \sec t,\ y = \tan t\)
This matches (I) \(\mathrm{cosec}\,t\).
\( \displaystyle \frac{dy}{dx} = \frac{\sec^{2} t}{\sec t\tan t} = \frac{\sec t}{\tan t} \)
\( \displaystyle = \frac{1}{\cos t} \cdot \frac{\cos t}{\sin t} = \mathrm{cosec}\,t \)