Ex 10.3, 20 - Chapter 10 Class 11 Conic Sections
Last updated at April 19, 2024 by Teachoo
Ex 10.3
Ex 10.3, 2 Important
Ex 10.3, 3
Ex 10.3, 4
Ex 10.3, 5 Important
Ex 10.3, 6
Ex 10.3, 7 Important
Ex 10.3, 8
Ex 10.3, 9
Ex 10.3, 10
Ex 10.3, 11 Important
Ex 10.3, 12 Important
Ex 10.3, 13
Ex 10.3, 14 Important
Ex 10.3, 15
Ex 10.3, 16 Important
Ex 10.3, 17
Ex 10.3, 18 Important
Ex 10.3, 19 Important
Ex 10.3, 20 You are here
Ex 10.3
Last updated at April 19, 2024 by Teachoo
Ex 10.3, 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 𝒙^𝟐/𝒂^𝟐 + 𝒚^𝟐/𝒃^𝟐 = 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)〗^2/𝑎^2 + 〖(3)〗^2/𝑏^2 = 1 16/𝑎^2 + 9/𝑏^2 = 1 Putting x = 6 & y = 2 in (1) 〖(6)〗^2/𝑎^2 + 〖(2)〗^2/𝑏^2 = 1 36/𝑎^2 + 4/𝑏^2 = 1 From (3) 16/𝑎^2 = 1 − 9/𝑏^2 1/𝑎^2 = 1/16 (1 − 9/𝑏^2 ) Putting value of 1/𝑎^2 in (2) 36/𝑎^2 + 4/𝑏^2 = 1 36(1/𝑎^2 ) + 4/𝑏^2 = 1 36(1/16 (1−9/𝑏^2 )) + 4/𝑏^2 = 1 36/16 (1−9/𝑏^2 ) + 4/𝑏^2 = 1 9/4 (1−9/𝑏^2 ) + 4/𝑏^2 = 1 9/4 − 81/〖4𝑏〗^2 + 4/𝑏^2 = 1 (−81)/(4𝑏^2 ) + 4/𝑏^2 = 1 − 9/4 (−81 + 16)/(4𝑏^2 ) = (4 − 9)/4 (−65)/(4𝑏^2 ) = (−5)/4 (−5)/4 (13/𝑏^2 )= (−5)/4 13/𝑏^2 = 1 1/𝑏^2 = 1/13 b2 = 13 Putting value of b2 in 1/𝑎^2 = 1/16 (1 − 9/𝑏^2 ) 1/𝑎^2 = 1/16 (1 − 9/13) 1/𝑎^2 = 1/16 ( (13 − 9)/13) 1/𝑎^2 = 1/16 ( 4/13) 1/𝑎^2 = 1/52a a2 = 52 Equation of ellipse is 𝑥^2/𝑎^2 + 𝑦^2/𝑏^2 = 1 Putting value of a2 & b2 𝒙^𝟐/𝟓𝟐 + 𝒚^𝟐/𝟏𝟑 = 1