Question

Find the locus of midpoints of points of intersection of the line $$ x \cos \theta + y \sin \theta = 1 $$ with the coordinate axes.

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

Hint:

Use parametric forms of intercepts and midpoint formula to derive the locus. Apply trigonometric identities for simplification.

Answer 1

The Correct Option is B

Given the line: \[ x \cos \theta + y \sin \theta = 1 \]
- X-intercept: put \( y = 0 \Rightarrow x = \frac{1}{\cos \theta} \)
- Y-intercept: put \( x = 0 \Rightarrow y = \frac{1}{\sin \theta} \) So intercepts are: \[ A = \left( \frac{1}{\cos \theta}, 0 \right),\quad B = \left( 0, \frac{1}{\sin \theta} \right) \] Midpoint \( M \) of \( AB \): \[ M = \left( \frac{1}{2\cos \theta}, \frac{1}{2\sin \theta} \right) \] Let the midpoint coordinates be \( (x, y) \), then: \[ x = \frac{1}{2\cos \theta} \Rightarrow \cos \theta = \frac{1}{2x}, \quad y = \frac{1}{2\sin \theta} \Rightarrow \sin \theta = \frac{1}{2y} \] Use identity: \[ \cos^2 \theta + \sin^2 \theta = 1 \Rightarrow \left(\frac{1}{2x}\right)^2 + \left(\frac{1}{2y}\right)^2 = 1 \Rightarrow \frac{1}{4x^2} + \frac{1}{4y^2} = 1 \Rightarrow \frac{1}{x^2} + \frac{1}{y^2} = 4 \Rightarrow \boxed{ \frac{1}{x^2} + \frac{1}{y^2} = \frac{1}{4} } \]