Question

The ends $A, B$ of a straight line segment of constant length $c$ slide upon the fixed rectangular axes $OX, OY$ respectively. If the rectangle $OAPB$ is completed, then the locus of the foot of the perpendicular drawn from $P$ to $AB$ is:

• A. $x^{2} + y^{2} = c^{2}$
• B. $x^{2/3} + y^{2/3} = c^{2/3}$
• C. $\sqrt{x} + \sqrt{y} = \sqrt{c}$
• D. $xy = c^{2}$

Hint:

This classic sliding rod problem is dynamically equivalent to finding the envelope of a ladder sliding down a wall. The paths traced out by the parametric perpendicular components always transition cleanly into fractional powers of the form $x^{2/3} + y^{2/3} = c^{2/3}$.

Answer 1

The Correct Option is B

Concept: Let the moving endpoints on the axes be represented as $A(a, 0)$ and $B(0, b)$. Since the segment has a fixed length of $c$, by the Pythagorean theorem, the parameters must always satisfy the constant structural constraint: $a^2 + b^2 = c^2$. Completing the rectangle fixes the coordinates of the opposite vertex at $P(a, b)$. Step 1: Write down the equations for lines $AB$ and $PF$.
Let the foot of the perpendicular drawn from $P(a,b)$ onto line segment $AB$ be defined as $F(x, y)$.
• The equation of the straight line $AB$ in intercept form is: \[ \frac{X}{a} + \frac{Y}{b} = 1 \quad \Rightarrow \quad bX + aY - ab = 0 \quad \text{(Slope } m_1 = -\frac{b}{a}\text{)} \]
• The line $PF$ passes through point $P(a,b)$ and is perpendicular to $AB$, meaning its slope is $m_2 = \frac{a}{b}$: \[ Y - b = \frac{a}{b}(X - a) \quad \Rightarrow \quad aX - bY - (a^2 - b^2) = 0 \]

Step 2: Solve for the coordinates of the intersection foot F.
The intersection point $F(x,y)$ satisfies both line equations. Solving this linear system for variables $x$ and $y$ yields: \[ x = \frac{a^3}{a^2 + b^2} \quad \text{and} \quad y = \frac{b^3}{a^2 + b^2} \] Substitute our constant parameter length constraint $a^2 + b^2 = c^2$ into the denominators: \[ x = \frac{a^3}{c^2} \quad \Rightarrow \quad a^3 = c^2 x \quad \Rightarrow \quad a = (c^2 x)^{\frac{1}{3}} \] \[ y = \frac{b^3}{c^2} \quad \Rightarrow \quad b^3 = c^2 y \quad \Rightarrow \quad b = (c^2 y)^{\frac{1}{3}} \]

Step 3: Substitute parameters into the length constraint equation.
Now substitute these isolated values for $a$ and $b$ back into our primary geometric constraint equation ($a^2 + b^2 = c^2$): \[ \left((c^2 x)^{\frac{1}{3}}\right)^2 + \left((c^2 y)^{\frac{1}{3}}\right)^2 = c^2 \quad \Rightarrow \quad c^{\frac{4}{3}} x^{\frac{2}{3}} + c^{\frac{4}{3}} y^{\frac{2}{3}} = c^2 \]

Step 4: Simplify exponents to find the final locus.
Divide the entire equation by the common exponential scaling term $c^{\frac{4}{3}}$: \[ x^{\frac{2}{3}} + y^{\frac{2}{3}} = \frac{c^2}{c^{\frac{4}{3}}} = c^{2 - \frac{4}{3}} = c^{\frac{2}{3}} \] This describes the standard mathematical locus for an astroid curve curve, matching option (B).