A normal is drawn at a point P(x, y) of a curve. It meets the x-axis at Q. If PQ is of constant leng

Q: Wait, why do we use capital $X$ and $Y$ for the normal line equation?

Great question! We use capital $X$ and $Y$ as variables for the line, while small $x$ and $y$ are the fixed coordinates of point $P$. It keeps our algebra clean and prevents us from confusing the variable coordinates of the line with the specific point $P$.

Q: Why does the $x$ term cancel out in the distance formula?

It is the magic of the geometry! The normal line's horizontal reach is determined by the slope. When you subtract the $x$-coordinate of $P$, you are left with just the 'run' of the normal, which is $y \frac{dy}{dx}$. It is a beautiful simplification that reveals the underlying structure.

Q: Is it always a circle?

Yes, for this specific condition. The constant length of the normal segment is a defining property of a circle. If the length were a function of $x$, we would get a different curve, but here, the geometry forces the curve to be a circle centered at the origin.

Question

A normal is drawn at a point $P(x, y)$ of a curve. It meets the x-axis at Q. If PQ is of constant length $k$, then show that the differential equation describing such curves is $y \frac{dy}{dx} = \pm \sqrt{k^2 - y^2}$. Find the equation of such a curve passing through (0, k).

Sigma Percentile · Accepted Answer

### Analyzing the Setup

My dear student, welcome to a beautiful exploration of geometry and calculus. Today, we are not just solving a problem; we are uncovering the hidden nature of a curve.

Imagine you are standing on a coordinate plane, looking at a curve defined by $y = f(x)$. You pick a point $P(x, y)$ on this curve. At this point, we draw a tangent, but our interest lies in the normal—the line perpendicular to that tangent.

This normal line stretches out until it hits the x-axis at a point $Q$. The problem tells us that the length of this segment $PQ$ is a constant $k$. This is our constraint, our anchor in the vast sea of possible curves.

### The Power of Slopes

To begin, we must translate this geometric description into the language of mathematics. The slope of the tangent at $P$ is $m = \frac{dy}{dx}$.

Because the normal is perpendicular to the tangent, its slope $m_N$ is the negative reciprocal:

$$m_N = -\frac{1}{dy/dx}$$

With this slope and the point $P(x, y)$, we can write the equation of the normal line using the point-slope form:

$$Y - y = -\frac{1}{dy/dx}(X - x)$$

Here, capital $X$ and $Y$ represent any point on the normal line, while small $x$ and $y$ are the fixed coordinates of our point $P$.

### Finding the X-Intercept

Now, we need to find the point $Q$ where this normal meets the x-axis. At any point on the x-axis, the y-coordinate is zero.

So, we set $Y = 0$ in our normal equation:

$$0 - y = -\frac{1}{dy/dx}(X - x)$$

Solving for $X$, we get $X = x + y \frac{dy}{dx}$. This gives us the coordinates of $Q$ as $(x + y \frac{dy}{dx}, 0)$. We have successfully pinned down point $Q$ in terms of the curve's position and its derivative.

### The Distance Constraint

The problem states that the length $PQ$ is a constant $k$. We use the distance formula:

$$PQ^2 = (X_Q - x_P)^2 + (Y_Q - y_P)^2 = k^2$$

Substituting our coordinates, we get:

$$(x + y \frac{dy}{dx} - x)^2 + (0 - y)^2 = k^2$$

Look at that! The $x$ terms cancel out, leaving us with:

$$\left(y \frac{dy}{dx}\right)^2 + y^2 = k^2$$

This is the differential equation that governs our curve. It is a beautiful, compact expression of the geometric constraint.

### Solving the Differential Equation

To find the curve, we must solve this equation. We isolate the derivative:

$$\left(y \frac{dy}{dx}\right)^2 = k^2 - y^2 \Rightarrow y \frac{dy}{dx} = \pm \sqrt{k^2 - y^2}$$

We separate the variables:

$$\frac{y}{\sqrt{k^2 - y^2}} dy = \pm dx$$

Now, we integrate both sides. The left side is a standard integral. By substituting $k^2 - y^2 = t^2$, we find the integral is $-\sqrt{k^2 - y^2}$. The right side is simply $\pm x + C$.

### The Final Revelation

We are given that the curve passes through $(0, k)$. Substituting these values, we find $C = 0$.

Our equation becomes:

$$-\sqrt{k^2 - y^2} = \pm x$$

Squaring both sides, we get $k^2 - y^2 = x^2$, or **$x^2 + y^2 = k^2$**.

This is the equation of a circle centered at the origin with radius $k$. We have traveled from a simple geometric constraint to the elegant, perfect symmetry of a circle. I hope you see the beauty in this journey!

Visualized Solution

Visualizing the Geometric Setup

Slopes of Tangent and Normal

Equation of the Normal Line

Finding the X-Intercept (Point $Q$ )

Solving for Coordinates of $Q$

Applying the Distance Formula

Substituting Coordinates into Distance Formula

The Differential Equation

Separating Variables for Integration

Integrating Both Sides

Executing the Integration

Finding the Constant of Integration

The Final Curve Equation

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Power of Slopes

Finding the X-Intercept

The Distance Constraint

Solving the Differential Equation

The Final Revelation

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