Question : If $x=\frac{\sqrt{5}+1}{\sqrt{5}-1}$ and $y=\frac{\sqrt{5}-1}{\sqrt{5}+1}$, then the value of $\frac{x^{2}+xy+y^{2}}{x^{2}-xy+y^{2}}$ is:
Option 1: $\frac{3}{4}$
Option 2: $\frac{4}{3}$
Option 3: $\frac{3}{5}$
Option 4: $\frac{5}{3}$
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Correct Answer: $\frac{4}{3}$
Solution :
Given: $x=\frac{\sqrt{5}+1}{\sqrt{5}-1}$ and $y=\frac{\sqrt{5}-1}{\sqrt{5}+1}$, then
$x^2=\frac{5+1+2\sqrt{5}}{5+1-2\sqrt{5}}$ and $y^2=\frac{5+1-2\sqrt{5}}{5+1+2\sqrt{5}}$
⇒ $x^2=\frac{6+2\sqrt{5}}{6-2\sqrt{5}}$ and $y^2=\frac{6-2\sqrt{5}}{6+2\sqrt{5}}$
and, $xy = \frac{\sqrt{5}+1}{\sqrt{5}-1}×\frac{\sqrt{5}-1}{\sqrt{5}+1}=1$
Putting the values, we get:
$\frac{x^{2}+xy+y^{2}}{x^{2}-xy+y^{2}}$
= $\frac{\frac{6+2\sqrt{5}}{6-2\sqrt{5}}+1+\frac{6-2\sqrt{5}}{6+2\sqrt{5}}}{\frac{6+2\sqrt{5}}{6-2\sqrt{5}}-1+\frac{6-2\sqrt{5}}{6+2\sqrt{5}}}$
= $\frac{\frac{(6+2\sqrt{5})^2+6^2-(2\sqrt5)^2+(6-2\sqrt{5})^2}{(6)^2-(2\sqrt{5})^2}}{\frac{(6+2\sqrt{5})^2-6^2+(2\sqrt5)^2+(6-2\sqrt{5})^2}{(6)^2-(2\sqrt{5})^2}}$
= $\frac{36+20+24\sqrt5+36-20+36+20-24\sqrt5}{36+20+24\sqrt5-36+20+36+20-24\sqrt5}$
= $\frac{128}{96}$ = $\frac{4}{3}$
Hence, the correct answer is $\frac{4}{3}$.
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