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JEE Advanced 2009Paper 1
Exam
JEE Advanced
Year
2009
Paper
Paper 1
Subject
Mathematics
Chapter
Hyperbola

JEE Advanced 2009 · Paper 1 · Mathematics · Hyperbola

Match the conics in Column I with the statements/expressions in...

 Match the conics in Column I with the statements/expressions in Column II. \text{ Match the conics in Column I with the statements/expressions in Column II. }Match the conics in Column I with the statements/expressions in Column II.
Match the conics in Column I with the statements/expressions in Column II.

Options

  1. A(A) s, (B) q,s, (C) s, (D) p,s
  2. B(A) p, (B) q,s, (C) r, (D) q,s
  3. C(A) s, (B) s,t, (C) s, (D) p,s
  4. D(A) p, (B) s,t, (C) r, (D) q,s

Correct answer

D. (A) p, (B) s,t, (C) r, (D) q,s

Step-by-step solution

(p) 1h2+k2=2⇒h2+k2=14\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∣∣=3z+2|-| z-2||=3 and 2−(−2)=4>3\quad 2-(-2)=4>3 Hence, locus is hyperbola. (r) Let x=3(1−t21+t2),y=2t1+t2x=\sqrt{3}\left(\frac{1-t^2}{1+t^2}\right), y=\frac{2 t}{1+t^2} Let tan⁡θ=t\quad \tan \theta=t ⇒x=3cos⁡2θ,y=sin⁡2θ\Rightarrow x=\sqrt{3} \cos 2 \theta, y=\sin 2 \theta ∴x23+y2=1\therefore \quad \frac{x^2}{3}+y^2=1 Hence, locus is an ellipse. (s) Eccentricity x=1⇒x=1 \Rightarrow Parabola, 1<x<∞⇒1 < x < \infty \Rightarrow Hyperbola (t) Let z=x+iyz=x+i y Since, Re⁡(z+1)2=∣z∣2+1\operatorname{Re}(z+1)^2=|z|^2+1 ⇒(x+1)2−y2=x2+y2+1\Rightarrow \quad(x+1)^2-y^2=x^2+y^2+1 ⇒2x=2y2\Rightarrow \quad 2 x=2 y^2 ⇒x=y2\Rightarrow \quad x=y^2 Hence, locus is parabola.
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