Match List-I with List-II

List-I	List-II
(Parametric equations)	$\left(\frac{dy}{dx}\right)$
(A) $x = \frac{2}{t}, y = 2t$	(I) $4t^2$
(B) $x = t^3, y = 3t + 2$	(II) $2(t+1)$
(C) $x = \log t, y = 2t^2$	(III) $-t^2$
(D) $x = e^t, y = 2te^t$	(IV) $t^{-2}$

Choose the correct answer from the options given below:

When x and y are functions of parameter t, the derivative is found using:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

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For $x = \frac{2}{t}, y = 2t$:

$x = 2t^{-1}$

$\frac{dx}{dt} = -2t^{-2}$

$= -\frac{2}{t^2}$

$\frac{dy}{dt} = 2$

$\frac{dy}{dx} = \frac{2}{-2/t^2}$

$= 2 \times \left(-\frac{t^2}{2}\right)$

$= -t^2$

(A) matches with (III)

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For $x = t^3, y = 3t + 2$:

$\frac{dx}{dt} = 3t^2$

$\frac{dy}{dt} = 3$

$\frac{dy}{dx} = \frac{3}{3t^2}$

$= \frac{1}{t^2}$

$= t^{-2}$

(B) matches with (IV)

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For $x = \log t, y = 2t^2$:

$\frac{dx}{dt} = \frac{1}{t}$

$\frac{dy}{dt} = 4t$

$\frac{dy}{dx} = \frac{4t}{1/t}$

$= 4t \times t$

$= 4t^2$

(C) matches with (I)

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For $x = e^t, y = 2te^t$:

$\frac{dx}{dt} = e^t$

Using product rule for $y = 2t \times e^t$:

$\frac{dy}{dt} = 2 \times e^t + 2t \times e^t$

$= 2e^t(1 + t)$

$\frac{dy}{dx} = \frac{2e^t(t + 1)}{e^t}$

$= 2(t + 1)$

(D) matches with (II)

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Final Matching:

(A) → (III): $-t^2$

(B) → (IV): $t^{-2}$

(C) → (I): $4t^2$

(D) → (II): $2(t+1)$