Match the conics in Column-I with the stateme | vedclass

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(A) Step $1$: Analyze Column-$II$ expressions.$P$: The line $hx + ky = 1$ touches $x^2 + y^2 = 4$ (radius $r=2$). The distance from origin to line is

Vedclass · Accepted Answer

$A 	o (P); B 	o (R, S); C 	o (Q, R)$

Vedclass · Answer

(A) Step $1$: Analyze Column-$II$ expressions.$P$: The line $hx + ky = 1$ touches $x^2 + y^2 = 4$ (radius $r=2$). The distance from origin to line is $1/\sqrt{h^2 + k^2} = 2 \implies h^2 + k^2 = 1/4$. This is a circle.$Q$: $|z + 2| - |z - 2| = \pm 3$. This represents a hyperbola with foci at $(\pm 2, 0)$ and $2a = 3 \implies a = 1.5$. Since $2ae = 4$,$e = 4/3 > 1$.$R$: Eccentricity $e \in [1, \infty)$ includes parabolas $(e=1)$ and hyperbolas $(e>1)$.$S$: $Re(z+1)^2 = |z|^2 + 1$. Let $z = x+iy$. $(x+1+iy)^2 = (x+1)^2 - y^2 + 2i(x+1)y$. $Re = (x+1)^2 - y^2$. Equation: $(x+1)^2 - y^2 = x^2 + y^2 + 1 \implies x^2 + 2x + 1 - y^2 = x^2 + y^2 + 1 \implies 2x = 2y^2 \implies x = y^2$. This is a parabola.Step $2$: Match with Column-$I$.$A$ (Circle) matches $P$.$B$ (Parabola) matches $S$ and $R$ (since $e=1$ for parabola).$C$ (Hyperbola) matches $Q$ and $R$ (since $e>1$ for hyperbola).Thus,$A 	o (P), B 	o (R, S), C 	o (Q, R)$.

Column-$I$	Column-$II$
$A$. Circle	$P$. Locus of point $(h, k)$ such that the line $hx + ky = 1$ touches the circle $x^2 + y^2 = 4$
$B$. Parabola	$Q$. Point $z$ in the complex plane satisfies $\|z + 2\| - \|z - 2\| = \pm 3$
$C$. Hyperbola	$R$. Eccentricity of the conic lies in the interval $[1, \infty)$
	$S$. Point $z$ in the complex plane satisfies $Re(z + 1)^2 = \|z\|^2 + 1$

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