The line joining `A(bcosalpha,bsinalpha)` and `B(acosbeta,asinbeta)` is produced to the point `M(x ,y)` so that `A M` and `B M` are in the ratio `b : adot` Then prove that `x+ytan((alpha+beta)/2)=0.`

Explanation

To solve the problem, we need to prove that \( x + y \tan\left(\frac{\alpha + \beta}{2}\right) = 0 \) given the points \( A(b \cos \alpha, b \sin \alpha) \) and \( B(a \cos \beta, a \sin \beta) \), and the point \( M(x, y) \) such that the segments \( AM \) and \( BM \) are in the ratio \( b : a \).

### Step-by-Step Solution:

1. **Identify the Coordinates of Points A and B**:
- Let \( A = (b \cos \alpha, b \sin \alpha) \)
- Let \( B = (a \cos \beta, a \sin \beta) \)

2. **Set Up the Ratio Condition**:
- The point \( M(x, y) \) divides the line segment \( AB \) in the ratio \( b : a \). Using the section formula, the coordinates of point \( M \) can be expressed as:
\[ M = \left( \frac{b \cdot (a \cos \beta) + a \cdot (b \cos \alpha)}{b + a}, \frac{b \cdot (a \sin \beta) + a \cdot (b \sin \alpha)}{b + a} \right) \]

3. **Calculate the Coordinates of M**:
- The x-coordinate of \( M \):
\[ x = \frac{b \cdot (a \cos \beta) + a \cdot (b \cos \alpha)}{b + a} \]
- The y-coordinate of \( M \):
\[ y = \frac{b \cdot (a \sin \beta) + a \cdot (b \sin \alpha)}{b + a} \]

4. **Express x and y in Terms of Sine and Cosine**:
- Factor out \( ab \):
\[ x = \frac{ab \left( \frac{b}{b + a} \cos \beta + \frac{a}{b + a} \cos \alpha \right)} \]
\[ y = \frac{ab \left( \frac{b}{b + a} \sin \beta + \frac{a}{b + a} \sin \alpha \right)} \]

5. **Divide x by y**:
- To find the relationship between \( x \) and \( y \):
\[ \frac{x}{y} = \frac{\frac{b \cos \beta + a \cos \alpha}{b + a}}{\frac{b \sin \beta + a \sin \alpha}{b + a}} = \frac{b \cos \beta + a \cos \alpha}{b \sin \beta + a \sin \alpha} \]

6. **Use Trigonometric Identities**:
- Using the identities for cosine and sine:
\[ \cos \beta - \cos \alpha = -2 \sin\left(\frac{\beta + \alpha}{2}\right) \sin\left(\frac{\beta - \alpha}{2}\right) \]
\[ \sin \beta - \sin \alpha = 2 \cos\left(\frac{\beta + \alpha}{2}\right) \sin\left(\frac{\beta - \alpha}{2}\right) \]

7. **Substituting Back**:
- Substitute these identities into the ratio:
\[ \frac{x}{y} = \frac{-2 \sin\left(\frac{\beta + \alpha}{2}\right) \sin\left(\frac{\beta - \alpha}{2}\right)}{2 \cos\left(\frac{\beta + \alpha}{2}\right) \sin\left(\frac{\beta - \alpha}{2}\right)} = -\tan\left(\frac{\alpha + \beta}{2}\right) \]

8. **Final Step**:
- Rearranging gives:
\[ x + y \tan\left(\frac{\alpha + \beta}{2}\right) = 0 \]

### Conclusion:
Thus, we have proved that \( x + y \tan\left(\frac{\alpha + \beta}{2}\right) = 0 \).