Question
If √(1−x6)+√(1−y6)=a(x3−y3) and dydx=f(x,y)√(1−y61−x6),then
Answer: Option D
:
D
Put x3=sin θ,y3=sin ϕ,
then cosθ+cosϕ=a(sin θ−sinϕ)
⇒2 cos(θ+ϕ2) cos(θ−ϕ2) =2a cos (θ+ϕ2) sin (θ−ϕ2)
⇒cot(θ−ϕ2) = a
⇒(θ−ϕ2)=cot−1a
⇒sin−1x3−sin−1y3=2 cot−1 a
∴3x2√(1−x6)−3y2√(1−y6)dydx=0
⇒dydx=x2y2√(1−y61−x6)
∴f(x,y)=x2y2
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:
D
Put x3=sin θ,y3=sin ϕ,
then cosθ+cosϕ=a(sin θ−sinϕ)
⇒2 cos(θ+ϕ2) cos(θ−ϕ2) =2a cos (θ+ϕ2) sin (θ−ϕ2)
⇒cot(θ−ϕ2) = a
⇒(θ−ϕ2)=cot−1a
⇒sin−1x3−sin−1y3=2 cot−1 a
∴3x2√(1−x6)−3y2√(1−y6)dydx=0
⇒dydx=x2y2√(1−y61−x6)
∴f(x,y)=x2y2
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