If the normal to the curve y = x + frac{2}{x} | vedclass

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(C) Given the curve $y = x + \frac{2}{x}$.At $x = 2$,$y = 2 + \frac{2}{2} = 3$. So,the point is $(2, 3)$.Differentiating with respect to $x$,we get $\

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$\frac{7\sqrt{5}}{2}$

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(C) Given the curve $y = x + \frac{2}{x}$.At $x = 2$,$y = 2 + \frac{2}{2} = 3$. So,the point is $(2, 3)$.Differentiating with respect to $x$,we get $\frac{dy}{dx} = 1 - \frac{2}{x^2}$.At $x = 2$,the slope of the tangent $m_t = 1 - \frac{2}{4} = 1 - \frac{1}{2} = \frac{1}{2}$.The slope of the normal $m_n = -\frac{1}{m_t} = -2$.The equation of the normal at $(2, 3)$ is $y - 3 = -2(x - 2)$,which simplifies to $y - 3 = -2x + 4$,or $2x + y = 7$.To find the points $A$ and $B$ where the normal meets the coordinate axes:For the $x$-intercept $(y = 0)$,$2x = 7 \implies x = \frac{7}{2}$. So,$A = (\frac{7}{2}, 0)$.For the $y$-intercept $(x = 0)$,$y = 7$. So,$B = (0, 7)$.The length of $AB = \sqrt{(\frac{7}{2} - 0)^2 + (0 - 7)^2} = \sqrt{\frac{49}{4} + 49} = \sqrt{49(\frac{1}{4} + 1)} = 7 \sqrt{\frac{5}{4}} = \frac{7\sqrt{5}}{2}$.

If the normal to the curve $y = x + \frac{2}{x}$ at the point where the abscissa is $2$ meets the coordinate axes at points $A$ and $B$,find the length of $AB$.

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