To determine the locus of the foot of the perpendicular from the origin to the line \(\frac{x}{a} + \frac{y}{b} = 1\), given the condition \(\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2}\), we follow these steps:
- Consider the equation of the line \(\frac{x}{a} + \frac{y}{b} = 1\). Rewriting it in the standard form: \(bx + ay = ab\).
- For a line in the form \(Ax + By = C\), the distance of the line from the origin is given by: \(d = \frac{|C|}{\sqrt{A^2 + B^2}}\). Thus, for our equation, \(A = b\), \(B = a\), and \(C = ab\), we have: \(d = \frac{|ab|}{\sqrt{a^2 + b^2}}\).
- The given condition is: \(\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2}\). Multiply through by \(a^2b^2c^2\) to get: \(b^2c^2 + a^2c^2 = a^2b^2\).
- Notice that: \(a^2b^2 = (ab)^2\) and \(b^2c^2 + a^2c^2 = c^2(a^2 + b^2)\).
Thus, we have: \(c^2(a^2 + b^2) = (ab)^2\). - Substitute back into the formula for the distance: \(d = \frac{|ab|}{\sqrt{a^2 + b^2}} = c\).
- This implies that the distance from the origin to the line is constant and equal to \(c\). The path traced by this distance from the origin is a circle with radius \(c\).
Thus, the locus of the foot of the perpendicular from the origin on the given line is a circle.