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

Options

A. `(2a)/(9)`
B. `(4a)/(9)`
C. `-(4a)/(9)`
D. `-(2a)/(9)`

Answer

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

### Step-by-step Solution:

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

2. **Finding the Slope of the Normal**: The normal to the curve at any point will have equal intercepts on the axes. If the intercepts are equal, the slope of the normal will be \( -1 \) (since the intercepts form a right triangle with equal sides).

3. **Finding the Slope of the Tangent**: The slope of the tangent line at any point on the curve can be found by differentiating the curve. We differentiate \( ay^2 = x^3 \) implicitly:
\[ \frac{d}{dx}(ay^2) = \frac{d}{dx}(x^3) \]
Using the chain rule, we get:
\[ 2ay \frac{dy}{dx} = 3x^2 \]
Thus, the slope of the tangent \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{3x^2}{2ay} \]

4. **Setting the Slope of the Tangent to 1**: Since the slope of the normal is \( -1 \), the slope of the tangent must be \( 1 \):
\[ \frac{3x^2}{2ay} = 1 \]

5. **Substituting \( y \)**: We know from the curve that \( y = \sqrt{\frac{x^3}{a}} \). Substitute this into the slope equation:
\[ \frac{3x^2}{2a\sqrt{\frac{x^3}{a}}} = 1 \]
Simplifying gives:
\[ \frac{3x^2}{2a \cdot \frac{x^{3/2}}{\sqrt{a}}} = 1 \implies \frac{3x^2 \sqrt{a}}{2ax^{3/2}} = 1 \]

6. **Cross Multiplying**: Cross multiplying gives:
\[ 3x^2 \sqrt{a} = 2ax^{3/2} \]

7. **Rearranging the Equation**: Rearranging gives:
\[ 3\sqrt{a} = 2ax^{-1/2} \]
Multiplying both sides by \( x^{1/2} \) gives:
\[ 3x^{1/2}\sqrt{a} = 2a \]

8. **Solving for \( x \)**: Squaring both sides results in:
\[ 9ax = 4a^2 \]
Dividing both sides by \( a \) (assuming \( a \neq 0 \)):
\[ 9x = 4a \implies x = \frac{4a}{9} \]

9. **Conclusion**: The abscissa of the point on the curve where the normal cuts off equal intercepts from the axes is:
\[ \boxed{\frac{4a}{9}} \]