Chapter 11 Class 11 Conic Sections
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Ex 10.4, 14 Find the equation of the hyperbola satisfying the given conditions: Vertices (Β±7, 0), e = 4/3 Here, the vertices are on the x-axis. Therefore, the equation of the hyperbola is of the form ππ/ππ β ππ/ππ = 1 Now, coor#dinates of vertices are (Β± a,0) & Given vertices = (Β±7, 0), So, (Β± a,0) = (Β±7, 0), a = 7 We know that Eccentricity = e = π/π Given that e = 4/3 4/3 = π/π 4a = 3c Putting a = 7 4 Γ 7=3 π 28 = 3 c 3c = 28 c = ππ/π Also, we know that c2 = a2 + b2 Putting values (28/3)^2 = 49 + b2 784/9 = 49 + b2 b2 = (784 β441)/9 b2 = πππ/π Required equation of hyperbola π₯2/π2β π¦2/π2 =1 Putting values π₯2/7^2 β π¦2/(343/9) =1 ππ/ππ β πππ/πππ = 1
