The minimum value of (x_1 - x_2)^2 + (sqrt{2 | vedclass

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(A) Let $y_1 = \sqrt{2 - x_1^2}$ and $y_2 = \frac{9}{x_2}$.Then $x_1^2 + y_1^2 = 2$ and $x_2 y_2 = 9$.Let $A = (x_1, y_1)$ and $B = (x_2, y_2)$. The e

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$8$

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(A) Let $y_1 = \sqrt{2 - x_1^2}$ and $y_2 = \frac{9}{x_2}$.Then $x_1^2 + y_1^2 = 2$ and $x_2 y_2 = 9$.Let $A = (x_1, y_1)$ and $B = (x_2, y_2)$. The expression represents the square of the distance $AB^2$ between point $A$ on the circle $x^2 + y^2 = 2$ and point $B$ on the hyperbola $xy = 9$.The minimum distance between two curves is along the common normal.The normal to the hyperbola $xy = 9$ at point $(3t, 3/t)$ is given by $y - 3/t = t^2(x - 3t)$.Since the circle is centered at the origin $(0, 0)$,the normal must pass through the origin.Substituting $(0, 0)$ into the normal equation: $-3/t = t^2(-3t)$ $\Rightarrow 3/t = 3t^3$ $\Rightarrow t^4 = 1$. Since $x_2 > 0$,we have $t = 1$.Point $B$ is $(3, 3)$ and point $A$ is the intersection of the line $y = x$ with the circle $x^2 + y^2 = 2$,which is $(1, 1)$.The distance $OB = \sqrt{3^2 + 3^2} = 3\sqrt{2}$ and $OA = \sqrt{1^2 + 1^2} = \sqrt{2}$.The minimum distance $AB = OB - OA = 3\sqrt{2} - \sqrt{2} = 2\sqrt{2}$.Therefore,$AB^2 = (2\sqrt{2})^2 = 8$.

The minimum value of $(x_1 - x_2)^2 + (\sqrt{2 - x_1^2} - \frac{9}{x_2})^2$ where $x_1 \in (0, \sqrt{2})$ and $x_2 \in R^+$.

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