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Forming Differential equations
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Forming Differential equations
Last updated at May 29, 2023 by Teachoo
Question 8 Form the differential equation of the family of ellipses having foci on π¦βππ₯ππ and center at origin. Equation of ellipse having center at origin (0, 0) & foci on y-axis is π₯^2/π^2 +π¦^2/π^2 =1 β΄ Differentiating Both Sides w.r.t. π₯ π/ππ₯ [π₯^2/π^2 +π¦^2/π^2 ] = (π(1))/ππ₯ 1/π^2 [2π₯]+1/π^2 [2π¦] ππ¦/ππ₯=0 2π₯/π^2 +2π¦/π^2 . ππ¦/ππ₯=0 Since it has two variables, we will differentiate twice 2π¦/π^2 π¦β²=(β2π₯)/π^2 π¦/π^2 π¦β²=(βπ₯)/π^2 (π¦/π₯)π¦β²=(βπ^2)/γ πγ^2 (π¦π¦^β²)/π₯ = (βπ^2)/π^2 Again differentiating both sides w.r.t. x ((π¦π¦^β² )^β² π₯ β (ππ₯/ππ₯)(π¦π¦^β² ))/π₯^2 =0 (π¦π¦^β² )^β² π₯ β (1)(π¦π¦^β² )=πΓπ^π (π¦π¦^β² )^β² π₯ βπ¦π¦^β²=π (Using Quotient rule and Diff. of constant is 0) (ππ^β² )^β² π₯ βπ¦π¦^β²=0 (π^β² π^β²+ππβ²β²)π₯ βπ¦π¦^β²=0 (γπ¦^β²γ^2+π¦π¦β²β²)π₯ βπ¦π¦^β²=0 π₯γπ¦^β²γ^2+π₯π¦π¦^β²β²βπ¦π¦^β²=0 πππ^β²β²+πγπ^β²γ^πβππ^β²=π (Using Product rule)