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 \(\frac{b}{a}=\)
- A \(\frac{1}{\sqrt{2} \mathrm{e}}\)
- B \(\frac{\mathrm{e}}{\sqrt{2}}\)
- C \(\sqrt{2} \mathrm{e}\)
- D \(\frac{\sqrt{2}}{\mathrm{e}}\)
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 \((\sqrt{2} \mathrm{e})\) : \((y-e)=m^{\prime}(x-\sqrt{2})\)…
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