Question 8 - Forming Differential equations - Chapter 9 Class 12 Differential Equations
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Question 8 Form the differential equation of the family of ellipses having foci on ๐ฆโ๐๐ฅ๐๐ and center at origin. Equation of ellipse having center at origin (0, 0) & foci on y-axis is ๐ฅ^2/๐^2 +๐ฆ^2/๐^2 =1 โด Differentiating Both Sides w.r.t. ๐ฅ ๐/๐๐ฅ [๐ฅ^2/๐^2 +๐ฆ^2/๐^2 ] = (๐(1))/๐๐ฅ 1/๐^2 [2๐ฅ]+1/๐^2 [2๐ฆ] ๐๐ฆ/๐๐ฅ=0 2๐ฅ/๐^2 +2๐ฆ/๐^2 . ๐๐ฆ/๐๐ฅ=0 Since it has two variables, we will differentiate twice 2๐ฆ/๐^2 ๐ฆโฒ=(โ2๐ฅ)/๐^2 ๐ฆ/๐^2 ๐ฆโฒ=(โ๐ฅ)/๐^2 (๐ฆ/๐ฅ)๐ฆโฒ=(โ๐^2)/ใ ๐ใ^2 (๐ฆ๐ฆ^โฒ)/๐ฅ = (โ๐^2)/๐^2 Again differentiating both sides w.r.t. x ((๐ฆ๐ฆ^โฒ )^โฒ ๐ฅ โ (๐๐ฅ/๐๐ฅ)(๐ฆ๐ฆ^โฒ ))/๐ฅ^2 =0 (๐ฆ๐ฆ^โฒ )^โฒ ๐ฅ โ (1)(๐ฆ๐ฆ^โฒ )=๐ร๐^๐ (๐ฆ๐ฆ^โฒ )^โฒ ๐ฅ โ๐ฆ๐ฆ^โฒ=๐ (Using Quotient rule and Diff. of constant is 0) (๐๐^โฒ )^โฒ ๐ฅ โ๐ฆ๐ฆ^โฒ=0 (๐^โฒ ๐^โฒ+๐๐โฒโฒ)๐ฅ โ๐ฆ๐ฆ^โฒ=0 (ใ๐ฆ^โฒใ^2+๐ฆ๐ฆโฒโฒ)๐ฅ โ๐ฆ๐ฆ^โฒ=0 ๐ฅใ๐ฆ^โฒใ^2+๐ฅ๐ฆ๐ฆ^โฒโฒโ๐ฆ๐ฆ^โฒ=0 ๐๐๐^โฒโฒ+๐ใ๐^โฒใ^๐โ๐๐^โฒ=๐ (Using Product rule)
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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 and Computer Science at Teachoo