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If $(3, \lambda \lambda \lambda \lambda)$ and $(5,6)$ are the conjugate points to the curve $x^{2}+y^{2}=3$, then $\lambda \lambda \lambda \lambda$ is
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$-2$
If the $(3, \lambda \lambda \lambda \lambda)$ and $(5,6)$ are the conjugate points to the curve $x^{2}+y^{2}=3$, then
$\begin{aligned} x_{1} x_{2}+y_{1} y_{2} &=3 \\ \Rightarrow \quad(3)(5)+(\lambda \lambda \lambda \lambda) \cdot(6) &=3 \end{aligned}$
$\Rightarrow \quad 15+6(\lambda \lambda \lambda \lambda)=3$
$\Rightarrow 6(\lambda \lambda \lambda \lambda)=-12$
$\Rightarrow \lambda \lambda \lambda \lambda=-2$
$\begin{aligned} x_{1} x_{2}+y_{1} y_{2} &=3 \\ \Rightarrow \quad(3)(5)+(\lambda \lambda \lambda \lambda) \cdot(6) &=3 \end{aligned}$
$\Rightarrow \quad 15+6(\lambda \lambda \lambda \lambda)=3$
$\Rightarrow 6(\lambda \lambda \lambda \lambda)=-12$
$\Rightarrow \lambda \lambda \lambda \lambda=-2$
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