The line $\sqrt{2} x + \sqrt{6} y = 2$ touches the hyperbola $x^2 - 2y^2 = 4$. The point of contact is:
Step-by-Step Solution
Key Concept: Condition for $lx + my = 1$ to touch $x^2/a^2 - y^2/b^2 = 1$: $a^2l^2 - b^2m^2 = 1$. Here line is $(\sqrt{2}/2)x + (\sqrt{6}/2)y = 1$. $a^2 = 4, b^2 = 2$: $4(1/2) - 2(3/2) = 2 - 3 = -1 \neq 1$. Rewrite: $\sqrt{2}x + \sqrt{6}y = 2$ touches $x^2/4 - y^2/2 = 1$. Point of contact formula: $(a^2l, -b^2m)/(1) = (4 \cdot \sqrt{2}/2, 2 \cdot \sqrt{6}/2)/(?)...$ Use $T = 0$ method.
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Correct Answer: (2)