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:
If a normal drawn at a point \( P \) to the curve \( y = \sin x \) passes through the origin, then the locus of \( P \) is:
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Use condition of normal passing through the origin to derive a locus equation.
Updated On: May 13, 2025
- \( x^2 = y^2 - y^4 \)
- \( x + y = 1 \)
- \( \frac{1}{y^2} + \frac{1}{x^2} = 1 \)
- \( \frac{1}{y^4} + \frac{1}{x^4} = 1 \)
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The Correct Option is A
Solution and Explanation
Given \( y = \sin x \), the slope of tangent = \( \cos x \) ⇒ slope of normal = \( -1/\cos x \)
Use normal line equation:
\[
y - \sin x = -\frac{1}{\cos x}(x - x) \Rightarrow \text{passes through origin ⇒ substitute (0,0)}
\Rightarrow -\sin x = \frac{x}{\cos x}
\Rightarrow x^2 = \sin^2 x - \sin^4 x \Rightarrow \boxed{x^2 = y^2 - y^4}
\]
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