Q107 / 160 · AP EAMCET 2023 (16 May Shift 2) · Mathematics

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

  • Correct answer
    A
    $\sqrt{3} \mathrm{x}+\mathrm{y}=\sqrt{3}$
  • B
    $x+\sqrt{3} y=\sqrt{3}$
  • C
    $3 x-\sqrt{3} y=0$
  • D
    $x+\sqrt{3} y=0$

Solution

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 $\begin{aligned} & y-\frac{\sqrt{3}}{2}=-\sqrt{3}\left(x-\frac{1}{2}\right) \\ & 2 \sqrt{3} x+2 y=2 \sqrt{3} \Rightarrow \sqrt{3} x+y=\sqrt{3} \end{aligned}$

Hence, the correct answer is option: 'A'.

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