web analytics

Example 13 - Find equation of ellipse, major axis along x-axis - Ellipse - Defination

Slide31.JPG
Slide32.JPG Slide33.JPG Slide34.JPG

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

Transcript

Example 13 Find the equation of the ellipse, with major axis along the x-axis and passing through the points (4, 3) and (– 1,4). Given that Major axis is along x-axis So required equation of ellipse is ﷐﷐𝒙﷮𝟐﷯﷮﷐𝒂﷮𝟐﷯﷯ + ﷐﷐𝒚﷮𝟐﷯﷮﷐𝒃﷮𝟐﷯﷯ = 1 Given that point (4, 3) & (−1, 4) lie of the ellipse So, point (4, 3) & (−1, 4) will satisfy equation of ellipse Putting x = 4 & y = 3 in (1) ﷐﷐𝑥﷮2﷯﷮﷐𝑎﷮2﷯﷯ + ﷐﷐𝑦﷮2﷯﷮﷐𝑏﷮2﷯﷯ = 1 ﷐﷐(4)﷮2﷯﷮﷐𝑎﷮2﷯﷯ + ﷐﷐(3)﷮2﷯﷮﷐𝑏﷮2﷯﷯ = 1 ﷐𝟏𝟔﷮﷐𝒂﷮𝟐﷯﷯ + ﷐𝟗﷮﷐𝒃﷮𝟐﷯﷯ = 1 Putting x = −1, y = 4 is in (1) ﷐﷐𝑥﷮2﷯﷮﷐𝑎﷮2﷯﷯ + ﷐﷐𝑦﷮2﷯﷮﷐𝑏﷮2﷯﷯ = 1 ﷐﷐(−1)﷮2﷯﷮﷐𝑎﷮2﷯﷯ + ﷐﷐(4)﷮2﷯﷮﷐𝑏﷮2﷯﷯ = 1 ﷐𝟏﷮﷐𝒂﷮𝟐﷯﷯ + ﷐𝟏𝟔﷮﷐𝒃﷮𝟐﷯﷯ = 1 Now, our equations are ﷐16﷮﷐𝑎﷮2﷯﷯ + ﷐9﷮﷐𝑏﷮2﷯﷯ = 1 …(2) ﷐1﷮﷐𝑎﷮2﷯﷯ + ﷐16﷮﷐𝑏﷮2﷯﷯ = 1 …(3) From (3) ﷐1﷮﷐𝑎﷮2﷯﷯+﷐16﷮﷐𝑏﷮2﷯﷯ = 1 ﷐1﷮﷐𝑎﷮2﷯﷯ = 1−﷐16﷮﷐𝑏﷮2﷯﷯ Putting value of ﷐1﷮﷐𝑎﷮2﷯﷯ in (2) ﷐16﷮﷐𝑎﷮2﷯﷯ + ﷐9﷮﷐𝑏﷮2﷯﷯ = 1 16﷐﷐1﷮﷐𝑎﷮2﷯﷯﷯ + ﷐9﷮﷐𝑏﷮2﷯﷯ = 1 16﷐1−﷐16﷮﷐𝑏﷮2﷯﷯﷯ + ﷐9﷮﷐𝑏﷮2﷯﷯ = 1 16 − ﷐256﷮﷐𝑏﷮2﷯﷯ + ﷐9﷮﷐𝑏﷮2﷯﷯ = 1 ﷐−256 + 9﷮﷐𝑏﷮2﷯﷯ = 1 −16 ﷐−247﷮﷐𝑏﷮2﷯﷯ = −15 b2 = ﷐−247﷮−15﷯ b2 = ﷐𝟐𝟒𝟕﷮𝟏𝟓﷯ Putting value of b2 = ﷐247﷮15﷯ in (3) ﷐1﷮﷐𝑎﷮2﷯﷯ = 1−﷐16﷮﷐𝑏﷮2﷯﷯ ﷐1﷮﷐𝑎﷮2﷯﷯ = 1−﷐16﷮﷐247﷮15﷯﷯ ﷐1﷮﷐𝑎﷮2﷯﷯ = 1−﷐16 × 15﷮247﷯ ﷐1﷮﷐𝑎﷮2﷯﷯ = ﷐247 − 240﷮247﷯ ﷐1﷮﷐𝑎﷮2﷯﷯ = ﷐7﷮247﷯ 𝐚𝟐 = ﷐𝟐𝟒𝟕﷮𝟕﷯ Thus, a2 = ﷐247﷮7﷯ & b2 = ﷐247﷮15﷯ Hence required of ellipse is ﷐﷐𝑥﷮2﷯﷮﷐𝑎﷮2﷯﷯ + ﷐﷐𝑦﷮2﷯﷮﷐𝑏﷮2﷯﷯ = 1 Putting values of a2 & b2 ﷐﷐𝑥﷮2﷯﷮﷐﷐247﷮7﷯﷯﷯ + ﷐﷐𝑦﷮2﷯﷮﷐﷐247﷮15﷯﷯﷯ = 1 ﷐7﷐𝑥﷮2﷯﷮247﷯ + ﷐15﷐𝑦﷮2﷯﷮247﷯ = 1 7x2 + 15y2 = 247

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 provides courses for Mathematics from Class 9 to 12. You can ask questions here.
Jail