Question

If the line \( \frac{x}{a} + \frac{y}{b} = 1 \) moves such that \( \frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2} \), then the locus of the foot of the perpendicular from the origin to the line is

• A. straight line
• B. circle
• C. parabola
• D. ellipse

Hint:

For moving lines with parameter conditions, convert locus using substitution carefully.

Answer 1

The Correct Option is B

Concept: Foot of perpendicular from origin to line \( Ax + By + C = 0 \) has coordinates: \[ \left( \frac{-AC}{A^2 + B^2}, \frac{-BC}{A^2 + B^2} \right) \]

Step 1: Convert line to standard form. \[ \frac{x}{a} + \frac{y}{b} - 1 = 0 \] \[ A = \frac{1}{a}, \quad B = \frac{1}{b}, \quad C = -1 \]

Step 2: Foot of perpendicular: \[ (x,y) = \left( \frac{\frac{1}{a}}{\frac{1}{a^2} + \frac{1}{b^2}}, \frac{\frac{1}{b}}{\frac{1}{a^2} + \frac{1}{b^2}} \right) \]

Step 3: Given: \[ \frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2} \] \[ (x,y) = (c^2 \cdot \tfrac{1}{a}, \; c^2 \cdot \tfrac{1}{b}) \]

Step 4: Then: \[ x^2 + y^2 = c^4\left( \frac{1}{a^2} + \frac{1}{b^2} \right) = c^4 \cdot \frac{1}{c^2} = c^2 \] Final Answer: \[ x^2 + y^2 = c^2 \Rightarrow \text{circle} \]