Ex 10.4

Chapter 10 Class 11 Conic Sections
Serial order wise

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### Transcript

Ex 10.4, 13 Find the equation of the hyperbola satisfying the given conditions: Foci (Β±4, 0), the latus rectum is of length 12 Since the foci are on the x-axis. Hence, the required equation of the hyperbola is ππ/ππ β ππ/ππ = 1 Now, coordinates of foci are (Β±c, 0) & given foci = (Β±4, 0) so, (Β±c,0) = (Β±4,0) c = 4 Now, Latus rectum =2π2/π Given latus rectum = 12 So, 2π2/π=12 2b2 = 12a b2 = 6a We know that c2 = a2 + b2 Putting value of c & b2 (4)2 = a2 + 6a 16 = a2 + 6a a2 + 6a β 16 = 0 a2 + 8a β 2a β16 = 0 a(a + 8) β 2 (a + 8) = 0 (a β 2) (a + 8) = 0 So, a = 2 or a = -8 Since βaβ is distance, it cannot be negative , So a = β8 is not possible β΄ a = 2, From (1) b2 = 6a b2 = 6 Γ2 b2 = 12 Thus, Required equation of hyperbola π₯2/π2 β π¦2/π2=1 Putting values π₯2/22 β π¦2/12=1 ππ/π β ππ/ππ=π