Ex 11.4, 12 - Find hyperbola: foci (35, 0), latus rectum 8 - Ex 11.4

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  1. Chapter 11 Class 11 Conic Sections
  2. Serial order wise
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Ex 11.4, 12 Find the equation of the hyperbola satisfying the given conditions: Foci (± 3﷐﷮5﷯, 0) , the latus rectum is of length 8. Co-ordinates of Foci is (±3﷐﷮5﷯, 0) Since foci is on the x-axis Hence equation of hyperbola is of the form ﷐x2﷮a2﷯ – ﷐y2﷮b2﷯ = 1 . Also, We know that co-ordinates of foci are (±c, 0) So, (±3﷐﷮5﷯, 0) = (±c, 0) ⇒ 3﷐﷮5﷯ = c ⇒ c = 3﷐﷮𝟓﷯ Now, c2 = a2 + b2 Putting c = 3﷐﷮5﷯ a2 + b2 = (3﷐﷮5﷯)2 a2 + b2 = 3 × 3 × ﷐﷮5﷯ × ﷐﷮5﷯ a2 + b2 = 9 × 5 a2 + b2 = 45 Also it is given that Latus Rectum = 8 ﷐2﷐𝑏﷮2﷯﷮𝑎﷯ = 8 2b2 = 8a b2 = ﷐8𝑎﷮2﷯ b2 = 4a Now, our equations are a2 + b2 = 45 …(1) b2 = 4a …(2) Putting the value of b2 in (1) a2 + 4a = 45 a2 + 4a − 45 = 0 a2 + 9a − 5a − 45 = 0 a(a + 9) − 5(a + 9) = 0 (a + 9) (a − 5) = 0 So, a = 5 or a = −9 Since a is distance, it can’t be negative So a = 5 From (2) b2 = 4a b2 = 4 × 5 b2 = 20 Equation of hyperbola is ﷐﷐𝑥﷮2﷯﷮﷐𝑎﷮2﷯﷯ − ﷐﷐𝑦﷮2﷯﷮﷐𝑏﷮2﷯﷯ = 1 ﷐﷐𝑥﷮2﷯﷮﷐5﷮2﷯﷯ − ﷐﷐𝑦﷮2﷯﷮20﷯ = 1 ﷐﷐𝒙﷮𝟐﷯﷮𝟐𝟓﷯ − ﷐﷐𝒚﷮𝟐﷯﷮𝟐𝟎﷯ = 1

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