Identification of Conic Sections from Loci and Parametric Forms

To determine the correct matches between Column I and Column II, we analyze each statement in Column II step-by-step:

1. Statement (p)

The line is given by $hx + ky = 1$ and it touches the circle $x^2 + y^2 = 4$ (radius $r = 2$, center $(0,0)$).
The perpendicular distance from the center $(0,0)$ to the line is equal to the radius:
$\frac{|h(0) + k(0) - 1|}{\sqrt{h^2 + k^2}} = 2 \implies \frac{1}{\sqrt{h^2 + k^2}} = 2 \implies h^2 + k^2 = \frac{1}{4}$
Replacing $(h, k)$ with $(x, y)$, the locus is:
$x^2 + y^2 = \frac{1}{4}$
This is the equation of a Circle.
Hence, (A) $\to$ (p).

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2. Statement (q)

The given condition in the complex plane is:
$|z + 2| - |z - 2| = \pm 3$
Let the two fixed points (foci) be $S_1(-2, 0)$ and $S_2(2, 0)$.
The distance between the foci is $2ae = 4$.
The locus represents points $P(z)$ such that the absolute difference between their distances to the two fixed points is constant:
$||z - (-2)| - |z - 2|| = 3$
Since $2a = 3 < 4 = 2ae$, we have the eccentricity $e = \frac{4}{3} > 1$.
This defines a Hyperbola.
Hence, (D) $\to$ (q).

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3. Statement (r)

The parametric equations are:
$x = \sqrt{3}\left(\frac{1-t^2}{1+t^2}\right), \quad y = \frac{2t}{1+t^2}$
Substituting $t = \tan\left(\frac{\theta}{2}\right)$, we use the standard half-angle identities $\cos\theta = \frac{1-t^2}{1+t^2}$ and $\sin\theta = \frac{2t}{1+t^2}$ to get:
$x = \sqrt{3}\cos\theta, \quad y = \sin\theta$
Eliminating the parameter $\theta$:
$\frac{x^2}{3} + \frac{y^2}{1} = \cos^2\theta + \sin^2\theta = 1$
This is the standard equation of an Ellipse.
Hence, (C) $\to$ (r).

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4. Statement (s)

The eccentricity $e$ of the conic lies in the interval:
$1 \le e < \infty$

• For a Parabola, the eccentricity is $e = 1 \in [1, \infty)$.
• For a Hyperbola, the eccentricity is $e > 1$, which also lies in $[1, \infty)$.
  Hence, (B) $\to$ (s) and (D) $\to$ (s).

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5. Statement (t)

Let $z = x + iy$. Then:
$z + 1 = (x + 1) + iy$
$(z + 1)^2 = ((x + 1) + iy)^2 = (x + 1)^2 - y^2 + 2i(x + 1)y$
Taking the real part:
$\text{Re}(z + 1)^2 = (x + 1)^2 - y^2 = x^2 + 2x + 1 - y^2$
The right-hand side is:
$|z|^2 + 1 = (x^2 + y^2) + 1$
Equating both sides:
$x^2 + 2x + 1 - y^2 = x^2 + y^2 + 1 \implies 2x = 2y^2 \implies y^2 = x$
This is the equation of a Parabola.
Hence, (B) $\to$ (t).

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Conclusion:

• (A) $\to$ (p)
• (B) $\to$ (s), (t)
• (C) $\to$ (r)
• (D) $\to$ (q), (s)

All the given options A, B, C, and D are correct.