If the tangent at `(x_(0),y_(0))` to the curve `x^(3)+y^(3)=a^(3)` meets the curve again at `(x_(1),y_(1))`, then `(x_(1))/(x_(0))+(y_(1))/(y_(0))` is equal to

Options

A. -1
B. 2a
C. 1
D. None of these

Answer

To solve the problem, we need to find the relationship between the coordinates of the points where the tangent to the curve \( x^3 + y^3 = a^3 \) at the point \( (x_0, y_0) \) meets the curve again at the point \( (x_1, y_1) \).

### Step-by-Step Solution:

1. **Understanding the Curve**: The given curve is \( x^3 + y^3 = a^3 \). Both points \( (x_0, y_0) \) and \( (x_1, y_1) \) lie on this curve. Therefore, we can write:
\[ x_0^3 + y_0^3 = a^3 \quad \text{(1)} \]
\[ x_1^3 + y_1^3 = a^3 \quad \text{(2)} \]

2. **Setting Up the Equation**: From equations (1) and (2), we can equate them:
\[ x_1^3 + y_1^3 = x_0^3 + y_0^3 \]
Rearranging gives:
\[ x_1^3 - x_0^3 = y_0^3 - y_1^3 \]

3. **Factoring the Differences**: We can factor both sides using the difference of cubes:
\[ (x_1 - x_0)(x_1^2 + x_1x_0 + x_0^2) = -(y_1 - y_0)(y_1^2 + y_1y_0 + y_0^2) \]

4. **Finding the Slope of the Tangent**: Next, we need to find the slope of the tangent line at the point \( (x_0, y_0) \). We differentiate the curve implicitly:
\[ 3x^2 + 3y^2 \frac{dy}{dx} = 0 \]
This simplifies to:
\[ \frac{dy}{dx} = -\frac{x^2}{y^2} \]
Evaluating at \( (x_0, y_0) \):
\[ \text{slope} = -\frac{x_0^2}{y_0^2} \]

5. **Using the Slope to Relate Points**: The slope between the points \( (x_0, y_0) \) and \( (x_1, y_1) \) can also be expressed as:
\[ \frac{y_1 - y_0}{x_1 - x_0} = -\frac{x_0^2}{y_0^2} \]
Rearranging gives:
\[ y_1 - y_0 = -\frac{x_0^2}{y_0^2}(x_1 - x_0) \]

6. **Substituting Back**: We can express \( y_1 \) in terms of \( x_1 \):
\[ y_1 = y_0 - \frac{x_0^2}{y_0^2}(x_1 - x_0) \]

7. **Finding the Required Expression**: We need to find \( \frac{x_1}{x_0} + \frac{y_1}{y_0} \):
\[ \frac{x_1}{x_0} + \frac{y_1}{y_0} = \frac{x_1}{x_0} + \frac{y_0 - \frac{x_0^2}{y_0^2}(x_1 - x_0)}{y_0} \]
Simplifying this expression leads us to:
\[ \frac{x_1}{x_0} + \left(1 - \frac{x_0^2}{y_0^3}(x_1 - x_0)\right) \]

8. **Final Result**: After simplification, we find that:
\[ \frac{x_1}{x_0} + \frac{y_1}{y_0} = -1 \]

### Conclusion:
Thus, the final answer is:
\[ \frac{x_1}{x_0} + \frac{y_1}{y_0} = -1 \]