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AP EAMCET · Maths · Application of Derivatives

The slope of the tangent at any point \((x, y)\) on the curve, is equal to the product of the coordinates of that point. If the equation of the normal to the curve at the point \((\sqrt{2}, e)\) is \(a x+b y=1\), then \(\frac{b}{a}=\)

  1. A \(\frac{1}{\sqrt{2} \mathrm{e}}\)
  2. B \(\frac{\mathrm{e}}{\sqrt{2}}\)
  3. C \(\sqrt{2} \mathrm{e}\)
  4. D \(\frac{\sqrt{2}}{\mathrm{e}}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\sqrt{2} \mathrm{e}\)

Step-by-step Solution

Detailed explanation

Slope of tangent at point \((\sqrt{2}, \mathrm{e})\) is \(\mathrm{m}=\mathrm{e} \sqrt{2}\) Then, slope of normal, \(m^{\prime}=-\frac{1}{m}=-\frac{1}{e \sqrt{2}}\) Equation of normal at point \((\sqrt{2} \mathrm{e})\) : \((y-e)=m^{\prime}(x-\sqrt{2})\)…