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$$
\text { Match the conics in Column I with the statements/expressions in Column II. }
$$
Options:
\text { Match the conics in Column I with the statements/expressions in Column II. }
$$
Solution:
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Verified Answer
The correct answer is:
(A) p, (B) s,t, (C) r, (D) q,s
(A) p, (B) s,t, (C) r, (D) q,s
(p) $\frac{1}{\sqrt{h^2+k^2}}=2 \Rightarrow h^2+k^2=\frac{1}{4}$
Hence, locus is a circle.
(q) ||$z+2|-| z-2||=3$
and $\quad 2-(-2)=4>3$
Hence, locus is hyperbola.
(r) Let $x=\sqrt{3}\left(\frac{1-t^2}{1+t^2}\right), y=\frac{2 t}{1+t^2}$
Let $\quad \tan \theta=t$ $\Rightarrow x=\sqrt{3} \cos 2 \theta, y=\sin 2 \theta$
$\therefore \quad \frac{x^2}{3}+y^2=1$
Hence, locus is an ellipse.
(s) Eccentricity $x=1 \Rightarrow$ Parabola, $1 < x < \infty \Rightarrow$ Hyperbola
(t) Let $z=x+i y$
Since, $\operatorname{Re}(z+1)^2=|z|^2+1$
$\Rightarrow \quad(x+1)^2-y^2=x^2+y^2+1$
$\Rightarrow \quad 2 x=2 y^2$
$\Rightarrow \quad x=y^2$
Hence, locus is parabola.
Hence, locus is a circle.
(q) ||$z+2|-| z-2||=3$
and $\quad 2-(-2)=4>3$
Hence, locus is hyperbola.
(r) Let $x=\sqrt{3}\left(\frac{1-t^2}{1+t^2}\right), y=\frac{2 t}{1+t^2}$
Let $\quad \tan \theta=t$ $\Rightarrow x=\sqrt{3} \cos 2 \theta, y=\sin 2 \theta$
$\therefore \quad \frac{x^2}{3}+y^2=1$
Hence, locus is an ellipse.
(s) Eccentricity $x=1 \Rightarrow$ Parabola, $1 < x < \infty \Rightarrow$ Hyperbola
(t) Let $z=x+i y$
Since, $\operatorname{Re}(z+1)^2=|z|^2+1$
$\Rightarrow \quad(x+1)^2-y^2=x^2+y^2+1$
$\Rightarrow \quad 2 x=2 y^2$
$\Rightarrow \quad x=y^2$
Hence, locus is parabola.
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