Ex 11.4, 12 - Chapter 11 Class 11 Conic Sections
Last updated at Feb. 6, 2020 by Teachoo
Last updated at Feb. 6, 2020 by Teachoo
Transcript
Ex 11.4, 12 Find the equation of the hyperbola satisfying the given conditions: Foci (Β± 3β5, 0) , the latus rectum is of length 8. Co-ordinates of Foci is (Β±3β5, 0) Since foci is on the x-axis Hence equation of hyperbola is of the form π₯2/π2 β π¦2/π2 = 1 . Also, We know that co-ordinates of foci are (Β±c, 0) So, (Β±3β5, 0) = (Β±c, 0) 3β5 = c c = 3βπ Now, c2 = a2 + b2 Putting c = 3β5 a2 + b2 = (3β5)2 a2 + b2 = 3 Γ 3 Γ β5 Γ β5 a2 + b2 = 9 Γ 5 a2 + b2 = 45 Also, it is given that Latus Rectum = 8 (2π^2)/π = 8 2b2 = 8a b2 = 8π/2 b2 = 4a Now, our equations are a2 + b2 = 45 b2 = 4a Putting the value of b2 in (1) a2 + b2 = 45 a2 + 4a = 45 a2 + 4a β 45 = 0 a2 + 9a β 5a β 45 = 0 a(a + 9) β 5(a + 9) = 0 (a + 9) (a β 5) = 0 So, a = 5 or a = β9 Since a is distance, it canβt be negative So a = 5 From (2) b2 = 4a b2 = 4 Γ 5 b2 = 20 Equation of hyperbola is π₯^2/π^2 β π¦^2/π^2 = 1 Putting values π₯^2/5^2 β π¦^2/20 = 1 π^π/ππ β π^π/ππ = 1
Ex 11.4
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