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Ex 9.3, 9 - Family of hyperbolas having foci on x-axis, center

Ex 9.3, 9 - Chapter 9 Class 12 Differential Equations - Part 2
Ex 9.3, 9 - Chapter 9 Class 12 Differential Equations - Part 3

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Question 9 Form the differential equation of the family of hyperbolas having foci on π‘₯βˆ’π‘Žπ‘₯𝑖𝑠 and center at origin. Equation of hyperbola having foci on x-axis & center at origin (0, 0) is π‘₯^2/π‘Ž^2 βˆ’π‘¦^2/𝑏^2 =1 ∴ Differentiating Both Sides w.r.t. π‘₯ 𝑑/𝑑π‘₯ [π‘₯^2/π‘Ž^2 βˆ’π‘¦^2/𝑏^2 ]=𝑑(1)/𝑑π‘₯ 1/π‘Ž^2 [2π‘₯]βˆ’1/𝑏^2 [2𝑦 . 𝑑𝑦/𝑑π‘₯]=0 2𝑦/𝑏^2 . 𝑦′=2π‘₯/π‘Ž^2 Since it has two variables, we will differentiate twice 𝑦/𝑏^2 𝑦′=π‘₯/π‘Ž^2 (𝑦/π‘₯)𝑦′=𝑏^2/π‘Ž^2 (𝑦𝑦^β€²)/π‘₯ = 𝑏^2/π‘Ž^2 Again differentiating both sides w.r.t. x ((𝑦𝑦^β€² )^β€² π‘₯ βˆ’ (𝑑π‘₯/𝑑π‘₯)(𝑦𝑦^β€² ))/π‘₯^2 =0 (𝑦𝑦^β€² )^β€² π‘₯ βˆ’ (1)(𝑦𝑦^β€² )=πŸŽΓ—π’™^𝟐 (𝑦𝑦^β€² )^β€² π‘₯ βˆ’π‘¦π‘¦^β€²=𝟎 (π’šπ’š^β€² )^β€² π‘₯ βˆ’π‘¦π‘¦^β€²=0 (Using Quotient rule and Diff. of constant is 0) (π’š^β€² π’š^β€²+π’šπ’šβ€²β€²)π‘₯ βˆ’π‘¦π‘¦^β€²=0 (〖𝑦^β€²γ€—^2+𝑦𝑦′′)π‘₯ βˆ’π‘¦π‘¦^β€²=0 π‘₯〖𝑦^β€²γ€—^2+π‘₯𝑦𝑦^β€²β€²βˆ’π‘¦π‘¦^β€²=0 π’™π’šπ’š^β€²β€²+π’™γ€–π’š^β€²γ€—^πŸβˆ’π’šπ’š^β€²=𝟎 (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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.