Ex 11.4, 11 - Find hyperbola: foci (0, 13), conjugate axis 24 - Hyperbola

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  1. Chapter 11 Class 11 Conic Sections
  2. Serial order wise
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Ex 11.4, 11 Find the equation of the hyperbola satisfying the given conditions: Foci (0, ±13), the conjugate axis is of length 24. We need to find equation of hyperbola given foci (0, ±13) & conjugate axis is of length 24. 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, ±13) So, (0, ± c) = (0, ±13) ⇒ c = 13 Length of conjugate axis = 2b Given length of conjugate axis = 24 So, 2b = 24 b = ﷐24﷮2﷯ b = 12 We know that c2 = a2 + b2 Putting Values (13)2= a2 + (12)2 (13)2 – (12)2 = a2 169 − 144 = a2 25 = a2 a2 = 25 Thus, the required equation of ellipse ﷐﷐𝑦﷮2﷯﷮﷐𝑎﷮2﷯﷯ − ﷐﷐𝑥﷮2﷯﷮﷐𝑏﷮2﷯﷯ =1 ﷐﷐𝑦﷮2﷯﷮25﷯ − ﷐﷐𝑥﷮2﷯﷮﷐12﷮2﷯﷯ =1 ﷐﷐𝒚﷮𝟐﷯﷮𝟐𝟓﷯ − ﷐﷐𝒙﷮𝟐﷯﷮𝟏𝟒𝟒﷯ =1

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