A curve passing through the point (1,2) and satisfying the condition that slope of the normal at any abscissa of that point , then the curve also passes through the point

Options

A. (0,0)
B. (2,2)
C. (2,1)
D. (3,2)

Answer

To solve the problem, we need to find the equation of the curve that passes through the point (1, 2) and has the property that the slope of the normal at any point on the curve is equal to the negative ratio of the coordinates of that point.

### Step-by-Step Solution:

1. **Understanding the Slope of the Normal**:
The slope of the normal to a curve at any point is given by \( -\frac{1}{\frac{dy}{dx}} \). According to the problem, this slope is equal to the negative ratio of the coordinates of the point, which can be expressed as:
\[ -\frac{1}{\frac{dy}{dx}} = \frac{y}{x} \]

2. **Setting Up the Equation**:
Rearranging the above equation gives:
\[ \frac{dy}{dx} = -\frac{x}{y} \]

3. **Separating Variables**:
We can separate the variables to integrate:
\[ y \, dy = -x \, dx \]

4. **Integrating Both Sides**:
Now we integrate both sides:
\[ \int y \, dy = \int -x \, dx \]
This results in:
\[ \frac{y^2}{2} = -\frac{x^2}{2} + C \]
Multiplying through by 2 to eliminate the fractions gives:
\[ y^2 + x^2 = 2C \]

5. **Finding the Constant of Integration**:
Since the curve passes through the point (1, 2), we can substitute \( x = 1 \) and \( y = 2 \) into the equation to find \( C \):
\[ 2^2 + 1^2 = 2C \implies 4 + 1 = 2C \implies 5 = 2C \implies C = \frac{5}{2} \]

6. **Final Equation of the Curve**:
Substituting \( C \) back into the equation gives:
\[ y^2 + x^2 = 5 \]

### Conclusion:
The equation of the curve is:
\[ x^2 + y^2 = 5 \]