Step 1: Understanding the Concept:
This is a locus problem involving a sliding rod of fixed length. We define the coordinates of the endpoints and the completed rectangle vertex, then find the condition for the foot of the perpendicular to derive the final locus equation.
Step 2: Key Formula or Approach:
1. Let \( A = (a, 0) \) and \( B = (0, b) \). Then \( a^2 + b^2 = c^2 \).
2. Vertex \( P = (a, b) \).
3. Line AB equation: \( \frac{x}{a} + \frac{y}{b} = 1 \).
4. Line PN (perpendicular from P to AB) equation.
Step 3: Detailed Explanation:
Let the foot of the perpendicular be \( N(h, k) \).
The slope of \( AB \) is \( m_1 = -b/a \).
Since \( PN \perp AB \), the slope of \( PN \) is \( m_2 = a/b \).
Equation of \( PN \): \( y - b = \frac{a}{b}(x - a) \implies bx - ay = b^2 - a^2 \).
Equation of \( AB \): \( bx + ay = ab \).
Solve for \( h, k \):
Add the equations: \( 2bx = ab + b^2 - a^2 \).
Wait, let's use a simpler method. \( N \) lies on AB, so \( bh + ak = ab \).
Also, the vector \( \vec{PN} \) is parallel to the normal of AB.
Standard locus theory for this configuration (foot of perpendicular from origin or rectangle vertex) frequently leads to the astroid form. Let's verify:
\( a = c \cos \theta, b = c \sin \theta \).
Equation of \( AB \): \( x \sin \theta + y \cos \theta = c \sin \theta \cos \theta \).
Slope of \( AB \) is \( -\tan \theta \). Slope of \( PN \) is \( \cot \theta \).
Equation of \( PN \): \( y - c \sin \theta = \cot \theta(x - c \cos \theta) \).
Simplifying and solving for \( x, y \), we obtain \( x = c \cos^3 \theta \) and \( y = c \sin^3 \theta \).
Eliminating \( \theta \): \( (x/c)^{2/3} + (y/c)^{2/3} = 1 \implies x^{2/3} + y^{2/3} = c^{2/3} \).
Step 4: Final Answer:
The locus is an astroid \( x^{2/3} + y^{2/3} = c^{2/3} \).