Question

The locus of the point on the curve \( y = \sin x \) where the tangent drawn at that point always passes through the point \( (0, \pi) \) is:

• A. \( x = y - \pi \)
• B. \( \sin x + \cos y + 1 = 0 \)
• C. \( x^2 (1 - y^2) = (y - \pi)^2 \)
• D. \( x^2 + (y - \pi)^2 = 0 \)

Hint:

Remember the equation of the tangent to a curve at a given point and use the condition that it passes through another point to derive the locus.

Answer 1

The Correct Option is C

Step 1: Find the equation of the tangent to the curve \( y = \sin x \) at a point \( (h, k) \) on the curve.
Since \( (h, k) \) lies on the curve \( y = \sin x \), we have \( k = \sin h \). The slope of the tangent at \( (h, k) \) is \( \frac{dy}{dx} \Big|_{x=h} = \cos h \). The equation of the tangent is \( y - k = \cos h (x - h) \).
Step 2: Use the condition that the tangent passes through the point \( (0, \pi) \).
Substituting \( (0, \pi) \) into the tangent equation: \[ \pi - k = \cos h (0 - h) \implies \pi - k = -h \cos h \]
Step 3: Eliminate \( h \) and \( k \) to find the locus in terms of \( x \) and \( y \).
We have \( k = \sin h \) and \( \pi - k = -h \cos h \). Replacing \( (h, k) \) with \( (x, y) \) for the locus: \[ y = \sin x \] \[ \pi - y = -x \cos x \] From the second equation, \( y - \pi = x \cos x \). Squaring both sides: \[ (y - \pi)^2 = x^2 \cos^2 x \] Using \( \cos^2 x = 1 - \sin^2 x = 1 - y^2 \): \[ (y - \pi)^2 = x^2 (1 - y^2) \] This matches option (3).