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

Slide46.JPG
Slide47.JPG Slide48.JPG Slide49.JPG

  1. Chapter 11 Class 11 Conic Sections
  2. Serial order wise
Ask Download

Transcript

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

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.