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The line $2 x+\sqrt{6} y=2$ is a tangent to the curve $x^{2}-2 y^{2}=4$. The point of contact is
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Verified Answer
The correct answer is:
$(4,-\sqrt{6})$
Solving the equation of line and curve, we get
$\begin{array}{l}
x^{2}-2\left\{\frac{2-2 x}{\sqrt{6}}\right\}^{2}=4 \\
\Rightarrow x^{2}-\frac{1}{3} \times 4\left(1+x^{2}-2 x\right)=4 \\
\Rightarrow 3 x^{2}-4-4 x^{2}+8 x=12 \\
\Rightarrow x^{2}-8 x+16=0 \\
\Rightarrow(x-4)^{2}=0 \Rightarrow x=4
\end{array}$
and $\sqrt{6} \cdot y=2-2(4)=-6$
$\Rightarrow y=-\sqrt{6}$
$\therefore$ Point of contact is $(4,-\sqrt{6})$.
$\begin{array}{l}
x^{2}-2\left\{\frac{2-2 x}{\sqrt{6}}\right\}^{2}=4 \\
\Rightarrow x^{2}-\frac{1}{3} \times 4\left(1+x^{2}-2 x\right)=4 \\
\Rightarrow 3 x^{2}-4-4 x^{2}+8 x=12 \\
\Rightarrow x^{2}-8 x+16=0 \\
\Rightarrow(x-4)^{2}=0 \Rightarrow x=4
\end{array}$
and $\sqrt{6} \cdot y=2-2(4)=-6$
$\Rightarrow y=-\sqrt{6}$
$\therefore$ Point of contact is $(4,-\sqrt{6})$.
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