Examples

Chapter 10 Class 11 Conic Sections
Serial order wise

### 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 1/𝑎^2 + 16/𝑏^2 = 1 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 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 (7𝑥^2)/247 + (15𝑦^2)/247 = 1 7x2 + 15y2 = 247

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.