See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

(I) Let z = x + iy ∴ 2 Im(Z ) 3 2xy 3    which is a rectangular hyperbola of eccentricity 2 . (II) Let ends of latus rectum are 2 b ae, a        and other focus is   ae,0  then 2 0 b / a tan30 2ae  2 2 2a e b 3   and   2 2 2 2 b a e 1 2e 3e 3 e 3        (III) AB = 6 = 2ae, PA – PB = 4 = 2a ∴ eccentricity of hyperbola is 3 e 2  Let eccentricity of conjugate hyperbola is e’ then 2 2 1 1 3 1 e' e (e') 5     (IV) Angle between asymptotes 1 2 2 2 2 2a 2ab b tan 3 3 a b a 1 b         Let eccentricity of conjugate hyperbola is e’ and a2   2 2 b (e') 1     2 2 2 e' 1 3 e' 2 (e') 2       For More Material Join: @JEEAdvanced_2024 AITS-CRT-III (Paper-1)-PCM(Sol.)-JEE(Advanced)/2023 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 11