Understanding Curve Area Division Properties

The question asks us to identify the type of curve that passes through the origin (0,0) and has a specific property related to area division within a rectangle formed by coordinate axes and lines parallel to them through a point (x, y) on the curve. The curve divides this rectangle into two regions, where one region's area is twice the other's.

Mathematical Formulation of the Area Division

Let's consider a point $(x, y)$ on the curve, where $x > 0$ and $y > 0$. The rectangle formed by the coordinate axes (x-axis and y-axis) and the lines $X=x$ and $Y=y$ has vertices at (0,0), (x,0), (x,y), and (0,y).

The total area of this rectangle is $A_{total} = x \times y$.

Let the curve be represented by the function $y = f(x)$. Since the curve passes through the origin, $f(0) = 0$. The area under the curve from $t=0$ to $t=x$ is given by the integral:

$$A_1 = \int_0^x f(t) dt$$

The curve divides the rectangle into two areas: $A_1$ (the area under the curve) and $A_2$ (the area above the curve within the rectangle).

$$A_2 = A_{total} - A_1 = xy - \int_0^x f(t) dt$$

The problem states that one area is twice the other. This gives two possible conditions:

• Condition 1: $A_1 = 2 A_2$
• Condition 2: $A_2 = 2 A_1$

Solving for the Curve Equation

Let's analyze these conditions:

Condition 1: $A_1 = 2 A_2$

$$A_1 = 2 (xy - A_1)$$ $$A_1 = 2xy - 2A_1$$ $$3A_1 = 2xy$$ $$A_1 = \frac{2}{3}xy$$

Substituting the integral form for $A_1$ and $y=f(x)$:

$$\int_0^x f(t) dt = \frac{2}{3} x f(x)$$

To find the function $f(x)$, we can differentiate both sides with respect to $x$. Using the Fundamental Theorem of Calculus on the left side:

$$f(x) = \frac{d}{dx} \left( \frac{2}{3} x f(x) \right)$$

Using the product rule for differentiation on the right side:

$$f(x) = \frac{2}{3} \left( 1 \cdot f(x) + x \cdot f'(x) \right)$$ $$f(x) = \frac{2}{3} f(x) + \frac{2}{3} x f'(x)$$

Rearranging the terms to form a differential equation:

$$f(x) - \frac{2}{3} f(x) = \frac{2}{3} x f'(x)$$ $$\frac{1}{3} f(x) = \frac{2}{3} x f'(x)$$ $$f(x) = 2x f'(x)$$

Let $y = f(x)$ and $f'(x) = \frac{dy}{dx}$. The equation becomes:

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

Separating variables:

$$\frac{dy}{y} = \frac{1}{2} \frac{dx}{x}$$

Integrating both sides:

$$\int \frac{dy}{y} = \int \frac{1}{2} \frac{dx}{x}$$ $$\ln|y| = \frac{1}{2} \ln|x| + C$$ $$\ln|y| = \ln|x^{1/2}| + C$$ $$y = e^C x^{1/2}$$

Let $A = e^C$. Then $y = A\sqrt{x}$. Squaring both sides gives $y^2 = A^2 x$. This equation represents a parabola opening along the positive x-axis.

Condition 2: $A_2 = 2 A_1$

$$xy - A_1 = 2 A_1$$ $$xy = 3A_1$$ $$A_1 = \frac{1}{3}xy$$

Substituting the integral form and $y=f(x)$:

$$\int_0^x f(t) dt = \frac{1}{3} x f(x)$$

Differentiating both sides with respect to $x$:

$$f(x) = \frac{d}{dx} \left( \frac{1}{3} x f(x) \right)$$ $$f(x) = \frac{1}{3} \left( 1 \cdot f(x) + x \cdot f'(x) \right)$$ $$f(x) = \frac{1}{3} f(x) + \frac{1}{3} x f'(x)$$

Rearranging the terms:

$$f(x) - \frac{1}{3} f(x) = \frac{1}{3} x f'(x)$$ $$\frac{2}{3} f(x) = \frac{1}{3} x f'(x)$$ $$2 f(x) = x f'(x)$$

Let $y = f(x)$ and $f'(x) = \frac{dy}{dx}$. The equation becomes:

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

Separating variables:

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

Integrating both sides:

$$\int \frac{dy}{y} = \int \frac{2dx}{x}$$ $$\ln|y| = 2 \ln|x| + C$$ $$\ln|y| = \ln|x^2| + C$$ $$y = e^C x^2$$

Let $A = e^C$. Then $y = Ax^2$. This equation represents a parabola opening along the positive y-axis (if $A>0$) or negative y-axis (if $A<0$).

Verification with a Parabola Example

Let's verify the second case, $y = Ax^2$. Here $f(x) = Ax^2$. The integral is $A_1 = \int_0^x At^2 dt = A \left[ \frac{t^3}{3} \right]_0^x = \frac{A x^3}{3}$. The total area of the rectangle is $xy = x(Ax^2) = Ax^3$. We need to check if $A_1 = \frac{1}{3}xy$. $$ \frac{Ax^3}{3} = \frac{1}{3} (Ax^3) $$ $$ \frac{Ax^3}{3} = \frac{Ax^3}{3} $$ This equality holds true. Therefore, a parabola satisfies the given conditions.

Conclusion

Both possible conditions for area division lead to equations of the form $y = Ax^2$ or $y = A\sqrt{x}$ (which is equivalent to $y^2 = Bx$). Both forms are equations of parabolas. Thus, the curve represents a family of parabolas.