A curve passes through `(2, 1)` and is such that the square of the ordinate is twice the rectangle contained by the abscissa and the intercept of the normal. Then the equation of curve is

Options

A. a. `x^(2)+y^(2)=9x`
B. b. `4x^(2)+y^(2)=9x`
C. c. `4x^(2)+2y^(2)=9x`
D. d. All of these

Answer

To solve the problem, we need to find the equation of a curve that passes through the point (2, 1) and satisfies the condition that the square of the ordinate (y) is twice the rectangle contained by the abscissa (x) and the intercept of the normal.

### Step-by-Step Solution:

1. **Equation of the Normal**:
The equation of the normal to a curve at a point (x, y) is given by:
\[ Y - y = -\frac{dx}{dy}(X - x) \]
Here, \( \frac{dx}{dy} \) is the derivative of the curve.

2. **Finding the y-intercept of the Normal**:
To find the intercept of the normal on the y-axis, we set \( Y = 0 \):
\[ 0 - y = -\frac{dx}{dy}(X - x) \]
Rearranging gives:
\[ y = \frac{dx}{dy}(X - x) \]
Setting \( X = 0 \) to find the x-intercept:
\[ x = y \frac{dy}{dx} \]

3. **Using the Given Condition**:
The problem states that the square of the ordinate is twice the rectangle formed by the abscissa and the intercept of the normal:
\[ y^2 = 2x \cdot \left(y \frac{dy}{dx}\right) \]
Simplifying this gives:
\[ y^2 = 2xy \frac{dy}{dx} \]

4. **Rearranging the Equation**:
Rearranging the equation leads to:
\[ \frac{dy}{dx} = \frac{y^2 - 2x^2}{2xy} \]

5. **Substituting \( y = vx \)**:
Let \( y = vx \), where \( v \) is a function of \( x \). Then:
\[ \frac{dy}{dx} = v + x \frac{dv}{dx} \]
Substituting this into the differential equation gives:
\[ v + x \frac{dv}{dx} = \frac{(vx)^2 - 2x^2}{2x(vx)} \]

6. **Simplifying**:
This simplifies to:
\[ v + x \frac{dv}{dx} = \frac{v^2x^2 - 2x^2}{2vx^2} \]
Further simplification leads to:
\[ 2v \cdot x \frac{dv}{dx} = -\frac{v^2 - 2}{2} \]

7. **Separating Variables**:
Rearranging gives:
\[ \frac{2v}{v^2 + 2} dv = -\frac{dx}{x} \]

8. **Integrating Both Sides**:
Integrating both sides:
\[ \int \frac{2v}{v^2 + 2} dv = -\int \frac{dx}{x} \]
The left side integrates to:
\[ \ln(v^2 + 2) \]
The right side integrates to:
\[ -\ln(x) + C \]

9. **Combining Results**:
This gives:
\[ \ln(v^2 + 2) = -\ln(x) + \ln(k) \]
Exponentiating both sides results in:
\[ v^2 + 2 = \frac{k}{x} \]

10. **Substituting Back for \( y \)**:
Since \( v = \frac{y}{x} \), we substitute back:
\[ \left(\frac{y}{x}\right)^2 + 2 = \frac{k}{x} \]
Multiplying through by \( x^2 \) gives:
\[ y^2 + 2x^2 = kx \]

11. **Using the Point (2, 1)**:
Substitute the point (2, 1) into the equation to find \( k \):
\[ 1^2 + 2 \cdot 2^2 = k \cdot 2 \]
This simplifies to:
\[ 1 + 8 = 2k \implies 9 = 2k \implies k = \frac{9}{2} \]

12. **Final Equation of the Curve**:
Substituting \( k \) back gives:
\[ y^2 + 2x^2 = \frac{9}{2} x \]
Rearranging yields:
\[ 2y^2 + 4x^2 = 9x \]

### Final Answer:
The equation of the curve is:
\[ 2y^2 + 4x^2 = 9x \]