Question

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

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

Hint:

Use condition of normal passing through the origin to derive a locus equation.

Answer 1

The Correct Option is A

Given \( y = \sin x \), the slope of tangent = \( \cos x \) ⇒ slope of normal = \( -1/\cos x \) Use normal line equation: \[ y - \sin x = -\frac{1}{\cos x}(x - x) \Rightarrow \text{passes through origin ⇒ substitute (0,0)} \Rightarrow -\sin x = \frac{x}{\cos x} \Rightarrow x^2 = \sin^2 x - \sin^4 x \Rightarrow \boxed{x^2 = y^2 - y^4} \]