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If the slope of the tangent on a curve at any point $(x, y)$ is equal to $\frac{y^2-x^2}{2 x y}$, then the equation of the normal at the point $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ is
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Verified Answer
The correct answer is:
$\sqrt{3} \mathrm{x}+\mathrm{y}=\sqrt{3}$
Given, $\frac{d y}{d x}=\frac{y^2-x^2}{2 x y}$
at, $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right) ; \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{3}{4}-\frac{1}{4}}{2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2}}=\frac{\frac{2}{4}}{\frac{2 \sqrt{3}}{4}}=\frac{1}{\sqrt{3}}$
Now, equation of normal is
$\begin{aligned}
& y-\frac{\sqrt{3}}{2}=-\sqrt{3}\left(x-\frac{1}{2}\right) \\
& 2 \sqrt{3} x+2 y=2 \sqrt{3} \Rightarrow \sqrt{3} x+y=\sqrt{3}
\end{aligned}$
at, $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right) ; \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{3}{4}-\frac{1}{4}}{2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2}}=\frac{\frac{2}{4}}{\frac{2 \sqrt{3}}{4}}=\frac{1}{\sqrt{3}}$
Now, equation of normal is
$\begin{aligned}
& y-\frac{\sqrt{3}}{2}=-\sqrt{3}\left(x-\frac{1}{2}\right) \\
& 2 \sqrt{3} x+2 y=2 \sqrt{3} \Rightarrow \sqrt{3} x+y=\sqrt{3}
\end{aligned}$
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