Question : If $ax+by=1$ and $bx+ay=\frac{2ab}{a^{2}+b^{2}}$, then $(x^{2}+y^{2})(a^{2}+b^{2})$ is equal to:
Option 1: 1
Option 2: 2
Option 3: 0.5
Option 4: 0
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Correct Answer: 1
Solution : $ax+by=1$ ---------------------------------------(1) $bx+ay=\frac{2ab}{a^{2}+b^{2}}$ -----------------------------------(2) On dividing equation (1) by (2), we get, ⇒ $\frac{ax+by}{bx+ay}=\frac{a^{2}+b^{2}}{2ab}$ ⇒ $2a^{2}xb+2ab^{2}y=a^{2}bx+b^{3}x+a^{3}y+ab^{2}y$ ⇒ $a^{2}xb+ab^{2}y=b^{3}x+a^{3}y$ ⇒ $a^{2}xb-b^{3}x=a^{3}y-ab^{2}y$ ⇒ $x(a^{2}b-b^{3})=y(a^{3}-ab^{2})$ ⇒ $\frac{x}{y}=\frac{(a^{3}-ab^{2})}{(a^{2}b-b^{3})}$ ⇒ $\frac{x}{y}=\frac{a(a^{2}-b^{2})}{b(a^{2}-b^{2})}$ ⇒ $\frac{x}{y}=\frac{a}{b}$ ---------------------------------(3) Putting the value of $(x,y)$ from equation (3) in equation (1), we get, ⇒ $y=\frac{b}{a^{2}+b^{2}}$ ⇒ $x=\frac{a}{a^{2}+b^{2}}$ ⇒ $x^{2}+y^{2}=\frac{a^{2}}{(a^{2}+b^{2})^{2}}+\frac{b^{2}}{(a^{2}+b^{2})^{2}}$ ⇒ $x^{2}+y^{2}=\frac{1}{(a^{2}+b^{2})}$ ⇒ $(x^{2}+y^{2})(a^{2}+b^{2})=1$ Hence, the correct answer is 1.
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