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Last updated at March 30, 2023 by Teachoo
Example 6 Form the differential equation representing the family of ellipses having foci on π₯βππ₯ππ is center at the origin. Ellipse whose foci is on x-axis & center at origin is π₯^2/π^2 +π¦^2/π^2 =1 Differentiating both sides w.r.t. π₯ π/ππ₯ [π₯^2/π^2 +π¦^2/π^2 ]=π(1)/ππ₯ 1/π^2 Γ(γπ(π₯γ^2))/ππ₯+1/π^2 Γ(γπ(π¦γ^2))/ππ₯=0 Since it has two variables, we will differentiate twice π₯^2/π^2 +π¦^2/π^2 =1 1/π^2 Γ2π₯+1/π^2 Γ(2π¦ . ππ¦/ππ₯)=0 2π₯/π^2 +2π¦/π^2 ππ¦/ππ₯=0 2π¦/π^2 ππ¦/ππ₯=(β2π₯)/γ πγ^2 π¦/π^2 ππ¦/ππ₯=(βπ₯)/γ πγ^2 π¦/π₯ ππ¦/ππ₯= (βπ^2)/γ πγ^2 π¦/π₯ π¦^β²= (βπ^2)/γ πγ^2 Again differentiating both sides π(π¦/π₯)/ππ₯. π¦^β²+π¦/π₯ (π(π¦^β²))/ππ₯=π/ππ₯ ((β π^2)/( π^2 )) [ππ¦/ππ₯ . π₯ β π¦ .ππ₯/ππ₯]/π₯^2 π¦^β² +π¦/π₯ Γπ¦β²β²=0 [π¦^β² π₯ β π¦]/π₯^2 π¦^β² +π¦/π₯Γπ¦β²β²=0 Multiplying x2 both sides π₯^2Γ[π¦^β² π₯ β π¦]/π₯^2 π¦^β² +π₯^2Γπ¦/π₯Γπ¦β²β²=π₯^2Γ0 [π¦^β² π₯βπ¦] π¦^β²+π₯π¦π¦^β²β²=0 γγπ₯π¦γ^β²γ^2βπ¦π¦^β²+π₯π¦π¦^β²β²=0 π₯π¦π¦^β²β²+γγπ₯π¦γ^β²γ^2βπ¦π¦^β²=0 ππ (π ^π π)/(π π^π ) +π(π π/π π)^πβπ π π/π π=π is the required differential equation