Question

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 \(\left(1, \frac{\sqrt{3}}{2}\right)\) is

• A. \(\sqrt{3}x + y = \sqrt{3}\)
• B. \(x + \sqrt{3}y = \sqrt{3}\)
• C. \(3x - \sqrt{3}y = 0\)
• D. \(x + \sqrt{3}y = 0\)

Hint:

For normals, use the negative reciprocal of the tangent's slope and apply the point-slope form.

Answer 1

The Correct Option is A

Given the slope of the tangent at any point is: \[ m = \frac{y^2 - x^2}{2xy} \] At \((1, \frac{\sqrt{3}}{2})\), substitute: \[ m = \frac{\left(\frac{\sqrt{3}}{2}\right)^2 - 1^2}{2 \cdot 1 \cdot \frac{\sqrt{3}}{2}} = \frac{\frac{3}{4} - 1}{\sqrt{3}} = \frac{-\frac{1}{4}}{\sqrt{3}} = -\frac{1}{4\sqrt{3}} \] So the slope of the tangent is \(-\frac{1}{4\sqrt{3}}\) The slope of the normal is the negative reciprocal: \[ m_n = 4\sqrt{3} \] Using point-slope form for the line: \[ y - \frac{\sqrt{3}}{2} = 4\sqrt{3}(x - 1) \Rightarrow y = 4\sqrt{3}x - 4\sqrt{3} + \frac{\sqrt{3}}{2} = 4\sqrt{3}x - \frac{7\sqrt{3}}{2} \] Multiply both sides by 2: \[ 2y = 8\sqrt{3}x - 7\sqrt{3} \Rightarrow 8\sqrt{3}x - 2y = 7\sqrt{3} \Rightarrow \sqrt{3}x + y = \sqrt{3} \]