The end A, B of a straight line segment of constant length c slide upon the fixed rectangular axes O

Q: Wait, why is $P$ at $(a, b)$?

Great question! Since $O$ is the origin $(0,0)$, $A$ is on the $x$-axis at $(a,0)$, and $B$ is on the $y$-axis at $(0,b)$, the rectangle $OAPB$ is formed by these points. The vertex $P$ opposite to the origin must have the coordinates $(a, b)$ to complete the rectangle.

Q: How do we know the slope of $PM$?

Remember that if two lines are perpendicular, the product of their slopes is $-1$. The line $AB$ has a slope of $-\frac{b}{a}$. Therefore, the slope of the perpendicular line $PM$ must be the negative reciprocal, which is $\frac{a}{b}$.

Q: Why do we eliminate $a$ and $b$?

The problem asks for the locus of the foot of the perpendicular. A locus is a path traced by a point, and its equation should only involve the coordinates $(x, y)$ of that point. Since $a$ and $b$ are just parameters describing the position of the ladder, they must be eliminated to find the final relationship between $x$ and $y$.

Question

The end $A, B$ of a straight line segment of constant length $c$ slide upon the fixed rectangular axes $OX, OY$ respectively. If the rectangle $OAPB$ be completed, then show that the locus of the foot of the perpendicular drawn from $P$ to $AB$ is $x^{2/3} + y^{2/3} = c^{2/3}$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a room, watching a ladder of constant length $c$ slide down a wall. The ladder touches the floor at point $A(a, 0)$ and the wall at point $B(0, b)$.

As it slides, the points $A$ and $B$ move, but the ladder itself remains rigid. This rigidity is our anchor. The distance between $A$ and $B$ is fixed, leading us to the fundamental constraint of our problem:

$$a^2 + b^2 = c^2$$

This equation is the heartbeat of our entire journey.

### The Rectangle and the Perpendicular

Now, let us complete the rectangle $OAPB$. With $O$ at the origin, $A$ at $(a, 0)$, and $B$ at $(0, b)$, the fourth vertex $P$ is naturally at $(a, b)$.

We are interested in the foot of the perpendicular $M$ dropped from $P$ onto the line $AB$. To find $M$, we first need the equation of the line $AB$. Using the intercept form, we have:

$$\frac{x}{a} + \frac{y}{b} = 1$$

This can be rearranged into the general form $bx + ay - ab = 0$.

Next, we consider the line $PM$. Since $PM$ is perpendicular to $AB$, its slope is the negative reciprocal of the slope of $AB$. The slope of $AB$ is $-\frac{b}{a}$, so the slope of $PM$ is $\frac{a}{b}$.

Using the point-slope form at $P(a, b)$, we get $y - b = \frac{a}{b}(x - a)$, which simplifies to:

$$ax - by = a^2 - b^2$$

### The Algebraic Dance

We now have a system of two linear equations: $bx + ay = ab$ and $ax - by = a^2 - b^2$. Our goal is to find the intersection point $M(x, y)$.

Multiplying the first equation by $b$ and the second by $a$, we get:

$$b^2x + aby = ab^2$$
$$a^2x - aby = a^3 - ab^2$$

Adding these two equations causes the $y$ terms to vanish, leaving us with $(a^2 + b^2)x = a^3$. Recalling our constraint $a^2 + b^2 = c^2$, we find:

$$x = \frac{a^3}{c^2}$$

By a symmetric process, we find:

$$y = \frac{b^3}{c^2}$$

### The Grand Unveiling

We have $x$ and $y$ in terms of $a$ and $b$. To find the locus, we must eliminate $a$ and $b$. From our expressions, we have $a = (xc^2)^{1/3}$ and $b = (yc^2)^{1/3}$.

Substituting these into our constraint $a^2 + b^2 = c^2$, we get:

$$(xc^2)^{2/3} + (yc^2)^{2/3} = c^2$$

Distributing the exponent, we have $x^{2/3}c^{4/3} + y^{2/3}c^{4/3} = c^2$. Dividing both sides by $c^{4/3}$, we arrive at the beautiful final equation:

$$x^{2/3} + y^{2/3} = c^{2/3}$$

This elegant curve is known as an **Astroid**, a star-like shape that emerges from the simple motion of a sliding ladder. You have just derived the equation of a classic geometric curve!

Visualized Solution

Visualizing the Setup

The Length Constraint

Completing the Rectangle $O A P B$

Equation of Line $A B$

Equation of Perpendicular $P M$

Finding Intersection Point $M$

Solving for $x$ -coordinate of $M$

Solving for $y$ -coordinate of $M$

Expressing Parameters $a$ and $b$

Substituting into the Constraint

Simplifying the Equation

The Final Locus Equation

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

The Rectangle and the Perpendicular

The Algebraic Dance

The Grand Unveiling

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Visualized Solution

Visualizing the Setup

The Length Constraint

Completing the Rectangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OAPB

Equation of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB

Equation of Perpendicular .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PM

Finding Intersection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x-coordinate of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y-coordinate of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M

Expressing Parameters .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b