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 š‘„āˆ’š‘Žš‘„š‘–š‘  and center at origin. Equation of hyperbola having foci on x-axis & center at origin (0, 0) is š‘„^2/š‘Ž^2 āˆ’š‘¦^2/š‘^2 =1 ∓ Differentiating Both Sides w.r.t. š‘„ š‘‘/š‘‘š‘„ [š‘„^2/š‘Ž^2 āˆ’š‘¦^2/š‘^2 ]=š‘‘(1)/š‘‘š‘„ 1/š‘Ž^2 [2š‘„]āˆ’1/š‘^2 [2š‘¦ . š‘‘š‘¦/š‘‘š‘„]=0 2š‘¦/š‘^2 . š‘¦ā€²=2š‘„/š‘Ž^2 Since it has two variables, we will differentiate twice š‘¦/š‘^2 š‘¦ā€²=š‘„/š‘Ž^2 (š‘¦/š‘„)š‘¦ā€²=š‘^2/š‘Ž^2 (š‘¦š‘¦^′)/š‘„ = š‘^2/š‘Ž^2 Again differentiating both sides w.r.t. x ((š‘¦š‘¦^′ )^′ š‘„ āˆ’ (š‘‘š‘„/š‘‘š‘„)(š‘¦š‘¦^′ ))/š‘„^2 =0 (š‘¦š‘¦^′ )^′ š‘„ āˆ’ (1)(š‘¦š‘¦^′ )=šŸŽĆ—š’™^šŸ (š‘¦š‘¦^′ )^′ š‘„ āˆ’š‘¦š‘¦^′=šŸŽ (š’šš’š^′ )^′ š‘„ āˆ’š‘¦š‘¦^′=0 (Using Quotient rule and Diff. of constant is 0) (š’š^′ š’š^′+š’šš’šā€²ā€²)š‘„ āˆ’š‘¦š‘¦^′=0 (ć€–š‘¦^′〗^2+š‘¦š‘¦ā€²ā€²)š‘„ āˆ’š‘¦š‘¦^′=0 š‘„ć€–š‘¦^′〗^2+š‘„š‘¦š‘¦^ā€²ā€²āˆ’š‘¦š‘¦^′=0 š’™š’šš’š^′′+š’™ć€–š’š^′〗^šŸāˆ’š’šš’š^′=šŸŽ (Using Product rule)

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