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(A) Given,the slope of the tangent is $\frac{dy}{dx} = \frac{y^2-x^2}{2xy}$.At the point $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$,the slope of t

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$\sqrt{3}x + y = \sqrt{3}$

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(A) Given,the slope of the tangent is $\frac{dy}{dx} = \frac{y^2-x^2}{2xy}$.At the point $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}ight)$,the slope of the tangent $m_t$ is:$m_t = \frac{(\frac{\sqrt{3}}{2})^2 - (\frac{1}{2})^2}{2(\frac{1}{2})(\frac{\sqrt{3}}{2})} = \frac{\frac{3}{4} - \frac{1}{4}}{\frac{\sqrt{3}}{2}} = \frac{\frac{2}{4}}{\frac{\sqrt{3}}{2}} = \frac{1}{2} 	imes \frac{2}{\sqrt{3}} = \frac{1}{\sqrt{3}}$.The slope of the normal $m_n$ is given by $-\frac{1}{m_t} = -\sqrt{3}$.The equation of the normal at $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}ight)$ is:$y - \frac{\sqrt{3}}{2} = -\sqrt{3}(x - \frac{1}{2})$$y - \frac{\sqrt{3}}{2} = -\sqrt{3}x + \frac{\sqrt{3}}{2}$$\sqrt{3}x + y = \sqrt{3}$.

If the slope of the tangent on a curve at any point $(x, y)$ is equal to $\frac{y^2-x^2}{2xy}$,then the equation of the normal at the point $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ is

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