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Hyperbola
Ex 10.4, 4
Ex 10.4, 2
Ex 10.4, 3 Important
Ex 10.4, 6
Ex 10.4, 5 Important
Example 14 (i)
Ex 10.4, 7 Important
Example 15
Ex 10.4, 8
Ex 10.4, 9 Important
Ex 10.4, 10
Ex 10.4, 11 Important
Ex 10.4, 12 Important
Ex 10.4, 13
Example 16 Important
Ex 10.4, 14 Important
Ex 10.4, 15 Important You are here
Last updated at May 29, 2023 by Teachoo
Ex 10.4, 15 Find the equation of the hyperbola satisfying the given conditions: Foci (0,±√10), passing through (2, 3) Since Foci is on the y−axis So required equation of hyperbola is 𝑦2/𝑎2 – 𝑥2/𝑏2 = 1 Now, Co-ordinates of foci = (0, ± c) & given foci = (0, ±√10) So, (0, ± c) = (0, ±√10) c = √𝟏𝟎 Also, c2 = a2 + b2 Putting value of c (√10)2 = a2 + b2 10 = a2 + b2 a2 + b2 = 10 b2 = 10 − a2 Since point (2,3) passes through the hyperbola, it satisfies the equation of hyperbola Putting values x = 2, y = 3 in equation 𝑦^2/𝑎^2 − 𝑥^2/𝑏^2 = 1 a 3^2/𝑎^2 − 2^2/𝑏^2 = 1 3^2/𝑎^2 − 2^2/𝑏^2 = 1 Also putting b2 = 10 − a2 3^2/𝑎^2 − 2^2/(10 − 𝑎^2 ) = 1 9/𝑎^2 − 4/(10 − 𝑎^2 ) = 1 ((10 − 𝑎^2 ) 9 − 4𝑎^2)/(𝑎^2 (10 − 𝑎^2)) = 1 (90 − 9𝑎^2 − 4𝑎^2)/(𝑎^2 (10 − 𝑎^2)) = 1 90 − 13a2 = a2(10 − a2) 90 − 13a2 = 10a2 − a4 a4 − 23a2 + 90 = 0 Let a2 = t So, our equation becomes t2 − 23t + 90 = 0 t2 − 18t − 5t + 90 = 0 t(t − 18) − 5(t − 18) = 0 (t − 5) (t − 18) = 0 So, t = 5 or t = 18 Now, a2 = t Thus, a2 = 5, 18 Putting t = 5 a2 = 5 Putting t = 18 a2 = 18 For a2 = 5 b2 = 10 – a2 b2 = 10 – 5 b2 = 5 For a2 = 18 b2 = 10 – a2 b2 = 10 – 18 b2 = –8 Since square cannot be negative a2 = 18 is not possible Thus, a2 = 5, b2 = 5 Required equation of hyperbola 𝑦^2/𝑎^2 − 𝑥^2/𝑏^2 =1 Putting values 𝒚^𝟐/𝟓 − 𝒙^𝟐/𝟓 =𝟏