The normal to the 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

Options

A. ellipse
B. parabola
C. circle
D. hyperbola or ellipse

Answer

To solve the problem step by step, we need to analyze the given information and derive the equations accordingly.

### Step 1: Understand the problem
We have a curve with a point \( P(x, y) \) on it. The normal to the curve at this point meets the x-axis at point \( G \). We are given that the distance from the origin to point \( G \) is twice the abscissa (x-coordinate) of point \( P \).

### Step 2: Write the equation of the normal
The equation of the normal to the curve at point \( P(x, y) \) is given by:
\[ y - y_1 = -\frac{dy}{dx}(x - x_1) \]
where \( (x_1, y_1) = (x, y) \).

### Step 3: Find the coordinates of point \( G \)
The normal meets the x-axis where \( y = 0 \). Setting \( y = 0 \) in the normal equation, we get:
\[ 0 - y = -\frac{dy}{dx}(x - x) \]
This simplifies to:
\[ G = \left(x + y \frac{dy}{dx}, 0\right) \]

### Step 4: Distance from the origin to point \( G \)
The distance from the origin to point \( G \) is given by the x-coordinate of \( G \):
\[ \text{Distance} = |x + y \frac{dy}{dx}| \]
According to the problem, this distance is twice the abscissa of point \( P \):
\[ |x + y \frac{dy}{dx}| = 2|x| \]

### Step 5: Set up the equation
From the equation above, we can write:
\[ x + y \frac{dy}{dx} = 2x \quad \text{or} \quad x + y \frac{dy}{dx} = -2x \]

### Step 6: Solve for \( y \frac{dy}{dx} \)
From \( x + y \frac{dy}{dx} = 2x \):
\[ y \frac{dy}{dx} = 2x - x = x \]
Thus, we have:
\[ y \frac{dy}{dx} = x \tag{1} \]

From \( x + y \frac{dy}{dx} = -2x \):
\[ y \frac{dy}{dx} = -2x - x = -3x \]
Thus, we have:
\[ y \frac{dy}{dx} = -3x \tag{2} \]

### Step 7: Separate variables and integrate
From equation (1):
\[ y \frac{dy}{dx} = x \implies y \, dy = x \, dx \]
Integrating both sides:
\[ \int y \, dy = \int x \, dx \implies \frac{y^2}{2} = \frac{x^2}{2} + C_1 \]
This simplifies to:
\[ y^2 = x^2 + 2C_1 \tag{3} \]

From equation (2):
\[ y \frac{dy}{dx} = -3x \implies y \, dy = -3x \, dx \]
Integrating both sides:
\[ \int y \, dy = \int -3x \, dx \implies \frac{y^2}{2} = -\frac{3x^2}{2} + C_2 \]
This simplifies to:
\[ y^2 + 3x^2 = 2C_2 \tag{4} \]

### Step 8: Identify the curves
From equation (3), we can rearrange it to:
\[ x^2 - y^2 = -2C_1 \]
This represents a hyperbola.

From equation (4), we can rearrange it to:
\[ 3x^2 + y^2 = 2C_2 \]
This represents an ellipse.

### Conclusion
Thus, the curve can be either a hyperbola or an ellipse depending on the constants \( C_1 \) and \( C_2 \).