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Ex 11.4, 15 - Find hyperbola: foci (0, 10), passing (2, 3) - Hyperbola

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  1. Chapter 11 Class 11 Conic Sections
  2. Serial order wise
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Ex 11.4, 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 ﷐﷐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 − 4﷮﷐𝑎﷮2﷯(10 − ﷐𝑎﷮2﷯)﷯ = 1 ﷐90 − 9﷐𝑎﷮2﷯ − 4﷐𝑎﷮2﷯﷮﷐𝑎﷮2﷯(10 − ﷐𝑎﷮2﷯)﷯ 90 − 13a2 = a2(10 − a2) 90 − 13a2 = 10a2 − a4 a4 − 10a2 − 13a2 + 90 = 0 a4 − 23a2 + 90 = 0 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 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

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