Suppose a curve whose sub tangent is n times the abscissa of the point of contact and passes through the point (2, 3). Then

Options

A. for n = 1, equation of the curve is `2y = 3x`
B. for n = 1, equation of the curve is `2y^(2) = 9x`
C. for n = 2, equation of the curve is `2y = 3x`
D. for n = 2, equation of the curve is `2y^(2) = 9x`

Explanation

To solve the problem, we need to find the equation of a curve whose sub-tangent is \( n \) times the abscissa of the point of contact and passes through the point \( (2, 3) \).

### Step-by-Step Solution:

1. **Understanding Sub-tangent**:
The sub-tangent of a curve at a point \( (x, y) \) is given by the formula:
\[ \text{Sub-tangent} = \frac{y \, dx}{dy} \]
According to the problem, the sub-tangent is \( n \) times the abscissa \( x \):
\[ \frac{y \, dx}{dy} = n x \]

2. **Rearranging the Equation**:
We can rearrange the equation to separate variables:
\[ y \, dx = n x \, dy \]
Dividing both sides by \( xy \):
\[ \frac{dx}{x} = \frac{n \, dy}{y} \]

3. **Integrating Both Sides**:
Now, we integrate both sides:
\[ \int \frac{dx}{x} = \int n \frac{dy}{y} \]
The integrals yield:
\[ \ln |x| = n \ln |y| + C \]

4. **Exponentiating**:
We can exponentiate both sides to eliminate the logarithm:
\[ |x| = e^C |y|^n \]
Let \( k = e^C \), so we have:
\[ x = k y^n \]

5. **Finding the Constant \( k \)**:
The curve passes through the point \( (2, 3) \). Thus, substituting \( x = 2 \) and \( y = 3 \):
\[ 2 = k (3^n) \]
Therefore, we can express \( k \) as:
\[ k = \frac{2}{3^n} \]

6. **Final Equation of the Curve**:
Substituting \( k \) back into the equation \( x = k y^n \):
\[ x = \frac{2}{3^n} y^n \]
Rearranging gives:
\[ 3^n x = 2 y^n \]
This can be rewritten as:
\[ 2 y^n = 3^n x \]

### Conclusion:
Thus, the equations of the curve for different values of \( n \) are:
- For \( n = 1 \):
\[ 2y = 3x \]
- For \( n = 2 \):
\[ 2y^2 = 9x \]