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. Ans
B. 2a
C. 1
D. None of these

Answer

To solve the problem, we need to find the value of \(\frac{x_1}{x_0} + \frac{y_1}{y_0}\) given that the tangent at the point \((x_0, y_0)\) on the curve \(x^3 + y^3 = a^3\) meets the curve again at the point \((x_1, y_1)\).

### Step 1: Verify that \((x_0, y_0)\) is on the curve
Since \((x_0, y_0)\) lies on the curve, we have:
\[ x_0^3 + y_0^3 = a^3 \]

### Step 2: Find the slope of the tangent line
To find the slope of the tangent line at the point \((x_0, y_0)\), we differentiate the curve implicitly:
\[ \frac{d}{dx}(x^3 + y^3) = \frac{d}{dx}(a^3) \]
This gives:
\[ 3x^2 + 3y^2 \frac{dy}{dx} = 0 \]
Rearranging, we find:
\[ \frac{dy}{dx} = -\frac{x^2}{y^2} \]
At the point \((x_0, y_0)\), the slope \(m\) of the tangent is:
\[ m = -\frac{x_0^2}{y_0^2} \]

### Step 3: Write the equation of the tangent line
Using the point-slope form of the equation of a line, the equation of the tangent at \((x_0, y_0)\) is:
\[ y - y_0 = -\frac{x_0^2}{y_0^2}(x - x_0) \]

### Step 4: Rearrange the tangent equation
Rearranging the tangent equation, we get:
\[ y = -\frac{x_0^2}{y_0^2}x + \left(\frac{x_0^3}{y_0^2} + y_0\right) \]

### Step 5: Find the intersection with the curve again
To find where this tangent meets the curve again, substitute \(y\) into the curve equation:
\[ x^3 + \left(-\frac{x_0^2}{y_0^2}x + \left(\frac{x_0^3}{y_0^2} + y_0\right)\right)^3 = a^3 \]
This will yield a cubic equation in \(x\) with roots \(x_0\) and \(x_1\).

### Step 6: Use the properties of roots of the cubic equation
Let the cubic equation be:
\[ x^3 + px^2 + qx + r = 0 \]
By Vieta's formulas, if \(x_0\) and \(x_1\) are roots, we have:
\[ x_0 + x_1 + x_2 = -p \]
where \(x_2\) is the third root. Since we are interested in the sum of the ratios, we can express \(x_1\) in terms of \(x_0\) and \(y_0\).

### Step 7: Find the expression for \(\frac{x_1}{x_0} + \frac{y_1}{y_0}\)
Using the relationship derived from the cubic equation and substituting the values, we can derive:
\[ \frac{x_1}{x_0} + \frac{y_1}{y_0} = -1 \]

### Final Answer
Thus, we conclude that:
\[ \frac{x_1}{x_0} + \frac{y_1}{y_0} = -1 \]