JEE Advanced · Mathematics · 15. Hyperbola
\[
\text { Match the conics in Column I with the statements/expressions in Column II. }
\]


- A
(A) s, (B) q,s, (C) s, (D) p,s
- B
(A) p, (B) q,s, (C) r, (D) q,s
- C
(A) s, (B) s,t, (C) s, (D) p,s
- D
(A) p, (B) s,t, (C) r, (D) q,s
Answer & Solution
Correct Answer
(D)
(A) p, (B) s,t, (C) r, (D) q,s
Step-by-step Solution
Detailed explanation
(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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