If \(x=a \sin t-b\) cost
\(y=a \operatorname{cost}+b \sin t\)
then \(y^{3} \frac{d^{2} y}{d x^{2}}+x^{2}+y^{2}=\)

(D)
We have, \(x=a \sin t-b \cos t\) and \(y=a \cos t+b \sin t\)
\(x^{2}+y^{2}=\left\{[a \sin t-b \cos t]^{2}+[a \cos t+b \sin t]^{2}\right\} \)
\( x^{2}+y^{2}=\{a^{2} \cdot \sin ^{2} t-2 a b \sin t \cdot \cos t+b^{2} \cos ^{2} t~+\) \(a^{2} \cos ^{2} t+2 a b \sin t \cdot \cos t+b^{2} \sin ^{2} t\} \)
\( x^{2}+y^{2}=\left\{a^{2}\left(\sin ^{2} t+\cos ^{2} t\right)+b^{2}\left(\cos ^{2} t+\sin ^{2} t\right)\right\} \)
\( x^{2}+y^{2}=a^{2}+b^{2}\)
Differentiate w.r.t. \(x\), we get
\(2 x+2 y \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=-\frac{x}{y}\)...(1)
\(\therefore \quad \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=-\mathrm{x}\)
Again differentiating w.r.t. x, we get
\(y \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}=-1\)
Substituting value of \(\frac{d y}{d x}\) from (1), we get
\( y \frac{d^{2} y}{d x^{2}}+\left(\frac{-x}{y}\right)^{2}=-1 \Rightarrow y \frac{d^{2} y}{d x^{2}}+\frac{x^{2}}{y^{2}}=-1 \)
\( \therefore y^{3} \frac{d^{2} y}{d x^{2}}+x^{2}+y^{2}=0\)