Match the following parametric forms in List- | vedclass

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(A) For $(A)$: Let $x = \frac{p}{2}\left(t+\frac{1}{t}\right)$ and $y = \frac{q}{2}\left(t-\frac{1}{t}\right)$.$\Rightarrow \frac{2x}{p} = t+\frac{1}{

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(A) For $(A)$: Let $x = \frac{p}{2}\left(t+\frac{1}{t}ight)$ and $y = \frac{q}{2}\left(t-\frac{1}{t}ight)$.$\Rightarrow \frac{2x}{p} = t+\frac{1}{t}$ and $\frac{2y}{q} = t-\frac{1}{t}$.Squaring and subtracting: $\left(\frac{2x}{p}ight)^2 - \left(\frac{2y}{q}ight)^2 = \left(t+\frac{1}{t}ight)^2 - \left(t-\frac{1}{t}ight)^2 = 4$.$\Rightarrow \frac{x^2}{p^2} - \frac{y^2}{q^2} = 1$,which represents a hyperbola. Thus,$(A ightarrow IV)$.For $(B)$: Let $x = p+q \cos 	heta$ and $y = r+q \sin 	heta$.$\Rightarrow (x-p) = q \cos 	heta$ and $(y-r) = q \sin 	heta$.Squaring and adding: $(x-p)^2 + (y-r)^2 = q^2(\cos^2 	heta + \sin^2 	heta) = q^2$.This represents a circle. Thus,$(B ightarrow II)$.For $(C)$: Let $x = p+\lambda^2$ and $y = q-\lambda$.$\Rightarrow \lambda = q-y$.Substituting $\lambda$ in $x$: $x = p + (q-y)^2$.$\Rightarrow (y-q)^2 = x-p$,which represents a parabola. Thus,$(C ightarrow I)$.Therefore,the correct match is $(A$ $ightarrow IV, B$ $ightarrow II, C$ $ightarrow I)$.

List-$I$	List-$II$
$(A)$ $\left[\frac{p}{2}\left(t+\frac{1}{t}\right), \frac{q}{2}\left(t-\frac{1}{t}\right)\right]$	$(I)$ parabola
$(B)$ $(p+q \cos \theta, r+q \sin \theta)$	$(II)$ circle
$(C)$ $(p+\lambda^2, q-\lambda)$	$(III)$ ellipse
	$(IV)$ hyperbola

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