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Example 15 - Find hyperbola foci (0,3), vertices (0, 11/2) - Hyperbola

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  1. Chapter 11 Class 11 Conic Sections
  2. Serial order wise
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Example 15 Find the equation of the hyperbola with foci (0, ± 3) and vertices ﷐0, ± ﷐﷐﷮11﷯﷮2﷯﷯. Since, foci are on the y-axis So required equation of hyperbola is ﷐𝒚𝟐﷮𝒂𝟐﷯ – ﷐𝒙𝟐﷮𝒃𝟐﷯ = 1 We know that Vertices = (0, ±a) Given vertices are ﷐0,± ﷐﷐﷮11﷯﷮2﷯﷯ So, (0, ±a) = ﷐0,± ﷐﷐﷮11﷯﷮2﷯﷯ a = ﷐﷐﷮11﷯﷮2﷯ a2 = ﷐𝟏𝟏﷮𝟒﷯ We know that foci = (0, ± c) Given foci = (0, ± 3) So c = 3 We know that c2 = a2 + b2 32 = ﷐11﷮4﷯ + b2 9 – ﷐11﷮4﷯ + b2 ﷐36−11﷮4﷯ = b2 ﷐25﷮4﷯ = b2 b2 = ﷐𝟐𝟓﷮𝟒﷯ Equation of hyperbola is ﷐﷐𝑦﷮2﷯﷮﷐𝑎﷮2﷯﷯ − ﷐﷐𝑥﷮2﷯﷮﷐𝑏﷮2﷯﷯ = 1 Putting values of a2 & b2 ﷐﷐𝑦﷮2﷯﷮﷐﷐11﷮4﷯﷯﷯ − ﷐﷐𝑥﷮2﷯﷮﷐﷐25﷮4﷯﷯﷯ = 1 ﷐4𝑦2﷮11﷯ − ﷐4𝑥2﷮25﷯ = 1 ﷐100𝑦2 − 44𝑥2﷮275﷯ = 1 100y2 − 44x2 = 275

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