Ex 11.4, 14 - Find hyperbola: vertices (7, 0), e = 4/3 - Ex 11.4

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Ex 11.4, 14 Find the equation of the hyperbola satisfying the given conditions: Vertices (±7, 0), e = ﷐4﷮3﷯ Here, the vertices are on the x-axis. Therefore, the equation of the hyperbola is of the form ﷐𝑥2﷮𝑎2﷯ – ﷐𝑦2﷮𝑏2﷯ = 1 . Now, coordinates of vertices are (± a,0) & given vertices = (±7, 0), So, (± a,0) = (±7, 0), a = 7 We know that Eccentricity = e = ﷐𝑐﷮𝑎﷯ Given that e = ﷐4﷮3﷯ ﷐4﷮3﷯ = ﷐𝑐﷮𝑎﷯ 4a = 3c Putting a = 7 4 × 7=3 𝑐 28 = 3 c 3c = 28 c = ﷐𝟐𝟖﷮𝟑﷯ Also, we know that c2 = a2 + b2 Putting values ﷐﷐﷐28﷮3﷯﷯﷮2﷯ = 49 + b2 ﷐784﷮9﷯ = 49 + b2 b2 = ﷐784 −441﷮9﷯ b2 = ﷐𝟑𝟒𝟑﷮𝟗﷯ Required equation of hyperbola ﷐𝑥2﷮𝑎2﷯ − ﷐𝑦2﷮𝑏2﷯ =1 Putting values ﷐𝑥2﷮49﷯ − ﷐𝑦2﷮﷐343﷮9﷯﷯ =1 ﷐𝒙𝟐﷮𝟒𝟗﷯ − ﷐𝟗𝒚𝟐﷮𝟑𝟒𝟑﷯ = 1

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