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AP EAMCET · Maths · Application of Derivatives

If a normal drawn at a point \(P\) to the curve \(y=\sin x\) passes through the origin, then the locus of \(\mathrm{P}\) is

  1. A \(\mathrm{x}^2=\mathrm{y}^2-\mathrm{y}^4\)
  2. B \(x+y=1\)
  3. C \(\frac{1}{y^2}-\frac{1}{x^2}=1\)
  4. D \(\frac{1}{y^4}-\frac{1}{x^4}=1\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\mathrm{x}^2=\mathrm{y}^2-\mathrm{y}^4\)

Step-by-step Solution

Detailed explanation

\(y=\sin x \Rightarrow \frac{d y}{d x}=\cos x\) Slope of normal \(\Rightarrow m=-\frac{1}{\frac{d y}{d x}}=-\frac{1}{\cos x}\) Let the co-ordinate of point \(\mathrm{P}\) is \((h, k)\) Then, \(m=-\frac{1}{\cos h}\) Equation of normal, which passes through \((0,0)\) is…