The minimum value of {left( {{x_1} - {x_2}} r | vedclass

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Question

$\mathrm{y}_{1}=\sqrt{2-\mathrm{x}_{1}^{2}}$ and $\mathrm{y}_{2}=\frac{9}{\mathrm{x}_{2}}$

$\Rightarrow \mathrm{x}_{1}^{2}+\mathrm{y}_{1}^{2}=2$ an

Vedclass · Accepted Answer

$8$

Vedclass · Answer

$\mathrm{y}_{1}=\sqrt{2-\mathrm{x}_{1}^{2}}$ and $\mathrm{y}_{2}=\frac{9}{\mathrm{x}_{2}}$

$\Rightarrow \mathrm{x}_{1}^{2}+\mathrm{y}_{1}^{2}=2$ and $\mathrm{x}_{2} \mathrm{y}_{2}=9.$

Let $A \equiv\left(x_{1}, y_{1}ight) B \equiv\left(x_{2}, y_{2}ight),$ then the given expression is $\mathrm{AB}^{2}$ where $\mathrm{A}$ and $\mathrm{B}$ are on curves $\mathrm{x}^{2}+\mathrm{y}^{2}=2$ and $\mathrm{xy}=9$ respectively for $AB$ to be minimum it must be a common normal.

Normal to $\mathrm{xy}=9$ at $\left(3 \mathrm{t}, \frac{3}{\mathrm{t}}ight)$

is $\mathrm{y}-\frac{3}{\mathrm{t}}=\mathrm{t}^{2}(\mathrm{x}-3 \mathrm{t})$ it will pass through $(0,0)$

$\Rightarrow-\frac{3}{t}=t^{2}(-3 t) \Rightarrow t=1$

$	herefore \mathrm{AB}_{\min }=3 \sqrt{2}-\sqrt{2}=2 \sqrt{2}$

$	herefore \mathrm{AB}^{2}=8$

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

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