# Ex 11.3, 20 - Chapter 11 Class 11 Conic Sections

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 11.3, 20 Find the equation for the ellipse that satisfies the given conditions: Major axis on the x-axis and passes through the points (4, 3) and (6, 2). Since Major axis is on the x-axis So required equation of ellipse is 𝒙𝟐𝒂𝟐 + 𝒚𝟐𝒃𝟐 = 1 Given that ellipse passes through point (4, 3) & (6, 2) Points (4, 3) & (6, 2) will satisfy the equation of ellipse Putting x = 4 & y = 3 in (1) (4)2𝑎2 + (3)2𝑏2 = 1 16𝑎2 + 9𝑏2 = 1 Putting x = 6 & y = 2 in (1) (6)2𝑎2 + (2)2𝑏2 = 1 36𝑎2 + 4𝑏2 = 1 Now equation are 16𝑎2 + 9𝑏2 = 1 …(2) 36𝑎2 + 4𝑏2 = 1 …(3) From (3) 16𝑎2 = 1 − 9𝑏2 1𝑎2 = 116 1 − 9𝑏2 Putting value of 1𝑎2 in (2) 36𝑎2 + 4𝑏2 = 1 361𝑎2 + 4𝑏2 = 1 361161−9𝑏2 + 4𝑏2 = 1 36161−9𝑏2 + 4𝑏2 = 1 941−9𝑏2 + 4𝑏2 = 1 94 − 81𝑏2 + 4𝑏2 = 1 −814𝑏2 + 4𝑏2 = 1 − 94 −81 + 164𝑏2 = 4 − 94 −654𝑏2 = −54 −54 13𝑏2= −54 13𝑏2 = 1 1𝑏2 = 113 b2 = 13 Putting value of b2 in 1𝑎2 = 1161 − 9𝑏2 1𝑎2 = 1161 − 913 1𝑎2 = 116 13 − 913 1𝑎2 = 116 413 1𝑎2 = 14 × 16 1𝑎2 = 152 a2 = 52 Equation of ellipse is 𝑥2𝑎2 + 𝑦2𝑏2 = 1 Putting value of a2 & b2 𝒙𝟐𝟓𝟐 + 𝒚𝟐𝟏𝟑 = 1

Chapter 11 Class 11 Conic Sections

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.