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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