Ex 11.4, 9 - Chapter 11 Class 11 Conic Sections (Term 2)
Last updated at Feb. 6, 2020 by Teachoo
Last updated at Feb. 6, 2020 by Teachoo
Transcript
Ex 11.4, 9 Find the equation of the hyperbola satisfying the given conditions: Vertices (0, Β±3), foci (0, Β±5) We need to find equation of hyperbola Given Vertices (0, Β±3), foci (0, Β±5) Since Vertices are on the y-axis So required equation of hyperbola is ππ/ππ β ππ/ππ = 1 β΄ Axis of hyperbola is yβaxis We know that Vertices = (0, Β±a) & Given vertices = (0, Β±3) β΄ (0, Β±a) = (0, Β±3) So a = 3 Also, coordinates of foci are (0, Β± c) Given, foci are (0, Β±5) So, c = 5 We know that c2 = a2 + b2 (5)2 = (3)2 + b2 25 = 9 + b2 b2 = 25 β 9 b2 = 16 Required equation of hyperbola is π¦2/π2 β π₯2/π2 = 1 Putting values π¦2/32 β π₯2/16 = 1 ππ/π β ππ/ππ = 1
Ex 11.4
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