Question
If x=secθ−cosθ,y=sec10θ−cos10θ and (x2+4)(dydx)2=k(y2+4), then k is equal to
Answer: Option D
:
D
∵x2+4=(secθ−cosθ)2+4=(secθ+cosθ)2⋯(i)Similarly,y2+4=(sec10θ+cos10θ)2⋯(ii)Now,dxdθ=secθtanθ+sinθ=tanθ(secθ+cosθ)anddydθ=10sec9θsecθtanθ−10cos9θ(−sinθ)=10tanθ(sec10θ−cos10θ)⇒dydx=(dydθ)(dxdθ)=10tanθ(sec10θ+cos10θ)tanθ(secθ+cosθ)∴(dydx)2=100(sec10θ+cos10θ)(secθ+cosθ)2=100(y2+4)(x2+4)or(x2+4)(dydx)2=100(y2+4)
On comparing with the expression given we get k = 100
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:
D
∵x2+4=(secθ−cosθ)2+4=(secθ+cosθ)2⋯(i)Similarly,y2+4=(sec10θ+cos10θ)2⋯(ii)Now,dxdθ=secθtanθ+sinθ=tanθ(secθ+cosθ)anddydθ=10sec9θsecθtanθ−10cos9θ(−sinθ)=10tanθ(sec10θ−cos10θ)⇒dydx=(dydθ)(dxdθ)=10tanθ(sec10θ+cos10θ)tanθ(secθ+cosθ)∴(dydx)2=100(sec10θ+cos10θ)(secθ+cosθ)2=100(y2+4)(x2+4)or(x2+4)(dydx)2=100(y2+4)
On comparing with the expression given we get k = 100
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