The abscissa of the point on the curve `ay^2 = x^3` , the normal at which cuts off equal intercepts from the axes is

Options

A. 2a/3
B. 4a/9
C. 3a
D. none

Answer

To solve the problem, we need to find the abscissa (x-coordinate) of the point on the curve \( ay^2 = x^3 \) where the normal at that point cuts off equal intercepts from the axes.

### Step-by-Step Solution:

1. **Understand the Curve**: The given curve is \( ay^2 = x^3 \). We can express \( y \) in terms of \( x \) as follows:
\[ y^2 = \frac{x^3}{a} \quad \Rightarrow \quad y = \pm \sqrt{\frac{x^3}{a}} \]

2. **Find the Slope of the Normal**: The normal to the curve at a point will cut off equal intercepts from the axes. Let the intercepts on the x-axis and y-axis be \( (p, 0) \) and \( (0, p) \) respectively. The slope of the line joining these two points is:
\[ \text{slope} = \frac{p - 0}{0 - p} = -1 \]
Therefore, the slope of the normal is \( -1 \).

3. **Find the Slope of the Tangent**: The slope of the tangent to the curve can be found using implicit differentiation. Differentiating \( ay^2 = x^3 \) with respect to \( x \):
\[ 2ay \frac{dy}{dx} = 3x^2 \quad \Rightarrow \quad \frac{dy}{dx} = \frac{3x^2}{2ay} \]

4. **Relate the Slopes**: Since the slope of the normal is the negative reciprocal of the slope of the tangent, we have:
\[ -\frac{3x^2}{2ay} = -1 \quad \Rightarrow \quad \frac{3x^2}{2ay} = 1 \]
Rearranging gives:
\[ 2ay = 3x^2 \quad \Rightarrow \quad y = \frac{3x^2}{2a} \]

5. **Substitute \( y \) back into the Curve**: Substitute \( y \) back into the original curve equation:
\[ a\left(\frac{3x^2}{2a}\right)^2 = x^3 \]
Simplifying this:
\[ a \cdot \frac{9x^4}{4a^2} = x^3 \quad \Rightarrow \quad \frac{9x^4}{4a} = x^3 \]

6. **Solve for \( x \)**: Cross-multiplying gives:
\[ 9x^4 = 4ax^3 \]
Dividing both sides by \( x^3 \) (assuming \( x \neq 0 \)):
\[ 9x = 4a \quad \Rightarrow \quad x = \frac{4a}{9} \]

### Final Answer:
The abscissa of the point on the curve \( ay^2 = x^3 \) at which the normal cuts off equal intercepts from the axes is:
\[ \boxed{\frac{4a}{9}} \]