Answer : Option B

Explanation :

$MF#%\begin{align}&x = \dfrac{\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)} = \dfrac{\left(\sqrt{3}+1\right)\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)} = \dfrac{\left(\sqrt{3}+1\right)^2}{3-1} = \dfrac{3 + 2\sqrt{3} + 1}{2}= \dfrac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}\\\\
&y = \dfrac{\sqrt{3}-1}{\sqrt{3}+1}= \dfrac{\left(\sqrt{3}-1\right)\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)} = \dfrac{\left(\sqrt{3}-1\right)^2}{3-1} = \dfrac{3 - 2\sqrt{3} + 1}{2}= \dfrac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}\\\\
&x^2 + y^2 = \left(2 + \sqrt{3}\right)^2 + \left(2 - \sqrt{3}\right)^2 = (4 + 4\sqrt{3}+3) + (4 - 4\sqrt{3}+3) = 2(4+3)= 14\end{align}$MF#%