The normal to a curve at P(x, y) meets the x-axis at G. If the distance of G from the origin is twic

Q: Wait, why is the slope of the normal $-\frac{dx}{dy}$ instead of $-\frac{dy}{dx}$?

Great catch! The slope of the tangent is $\frac{dy}{dx}$. Since the normal is perpendicular to the tangent, its slope must be the negative reciprocal of the tangent's slope. So, it is $-\frac{1}{dy/dx}$, which is indeed $-\frac{dx}{dy}$. This is a crucial distinction when working with differential equations.

Q: Why do we set $Y=0$ to find the x-intercept?

Because the problem states that the normal meets the x-axis at $G$. Any point on the x-axis must have a y-coordinate of zero. By setting $Y=0$ in the equation of the normal, we are mathematically forcing the line to intersect the x-axis, allowing us to find the corresponding X-coordinate.

Q: Is the constant $C$ important? Does it change the shape of the curve?

The constant $C$ is essential for the general solution of the differential equation. It represents a family of curves. In this case, $x^2 - y^2 = K$ (where $K = -2C$) defines a family of rectangular hyperbolas. The value of $K$ determines the specific size and position of the hyperbola, but the fundamental shape remains the same.

Question

The normal to a curve at $P(x, y)$ meets the x-axis at $G$. If the distance of $G$ from the origin is twice the abscissa of $P$, then the curve is a

Options:

(A) circle

(B) hyperbola

(C) ellipse

(D) parabola.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a smooth, winding path represented by a curve on a coordinate plane. You pick a point $P(x, y)$ on this path. At this exact spot, you draw a line perpendicular to the path—this is the normal.

As you follow this line, it eventually crosses the x-axis at a point $G$. The problem gives us a fascinating geometric constraint: the distance from the origin $O$ to this point $G$ is exactly twice the abscissa (the x-coordinate) of your starting point $P$.

Let us translate this geometric story into the language of calculus.

### Translating Geometry to Algebra

To find the coordinates of $G$, we first need the equation of the normal line. We know the slope of the tangent at $P$ is $\frac{dy}{dx}$. Therefore, the slope of the normal is its negative reciprocal, $-\frac{dx}{dy}$.

Using the point-slope form of a line, the equation of the normal at $P(x, y)$ is:

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

Here, $(X, Y)$ represents any point on the normal line. Since $G$ lies on the x-axis, its coordinates must be $(X, 0)$. By substituting $Y = 0$ into our equation, we can isolate $X$:

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

Multiplying both sides by $-\frac{dy}{dx}$, we get:

$$y \frac{dy}{dx} = X - x$$

Rearranging this, we find the x-coordinate of $G$:

$$X = x + y \frac{dy}{dx}$$

### The Differential Equation

The problem states that the distance $OG$ is twice the abscissa of $P$. Since $G$ is on the x-axis, the distance $OG$ is simply the absolute value of its x-coordinate, $X$.

Thus, we have the condition $X = 2x$. Substituting our expression for $X$ into this condition, we get:

$$x + y \frac{dy}{dx} = 2x$$

Subtracting $x$ from both sides, we arrive at a beautiful, simple differential equation:

$$y \frac{dy}{dx} = x$$

### The Integration

Now, we enter the realm of integration. To solve this, we separate the variables by moving all $y$ terms to one side and all $x$ terms to the other:

$$y \, dy = x \, dx$$

Integrating both sides, we get:

$$\int y \, dy = \int x \, dx$$

$$\frac{y^2}{2} = \frac{x^2}{2} + C$$

Multiplying by $2$ to clear the denominators, we have $y^2 = x^2 + 2C$. Rearranging the terms, we get:

$$x^2 - y^2 = -2C$$

If we let $-2C = K$, we get **$x^2 - y^2 = K$**. This is the standard equation of a **rectangular hyperbola**.

It is truly elegant how the simple constraint of the normal's intersection leads us directly to one of the fundamental conic sections!

Animated Solution for Mathematics - Differential Equations: The normal to a curve at $P (x, y)$ meets the x-axis at $G$ . If the distance of $G$ from the origin is twice the abscissa of $P$ , then the curve is a

Select Answer:

Visualized Solution

Visualizing the Curve and Point $P$

Drawing the Normal

Equation of the Normal

Finding Coordinates of $G$

Solving for $X$ -coordinate of $G$

The Abscissa of $P$

Applying the Given Condition

Setting up the Equation

Simplifying the Differential Equation

Separating the Variables

Integrating Both Sides

Identifying the Curve

The Sigma Insight: Variable Separable Method

Analyzing the Setup

Translating Geometry to Algebra

The Differential Equation

The Integration

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