The equation `x^(2)+2xy+4=0` is transformed to the parallel axes through the point `(6, lambda)`. For what value of `lambda` its new form passes through the new origin ?

Answer

To solve the problem, we need to find the value of \( \lambda \) such that the transformed equation \( x^2 + 2xy + 4 = 0 \) passes through the new origin at the point \( (6, \lambda) \).

### Step-by-Step Solution:

1. **Write down the given equation**:
The equation provided is:
\[ x^2 + 2xy + 4 = 0 \]

2. **Substitute the new origin**:
We need to check if the point \( (6, \lambda) \) satisfies the equation. Substitute \( x = 6 \) and \( y = \lambda \) into the equation:
\[ 6^2 + 2(6)(\lambda) + 4 = 0 \]

3. **Calculate \( 6^2 \) and simplify**:
Calculate \( 6^2 \):
\[ 6^2 = 36 \]
Now substitute this back into the equation:
\[ 36 + 12\lambda + 4 = 0 \]

4. **Combine like terms**:
Combine \( 36 \) and \( 4 \):
\[ 40 + 12\lambda = 0 \]

5. **Isolate \( \lambda \)**:
Rearranging the equation gives:
\[ 12\lambda = -40 \]
Now divide both sides by \( 12 \):
\[ \lambda = -\frac{40}{12} \]

6. **Simplify the fraction**:
Simplify \( -\frac{40}{12} \):
\[ \lambda = -\frac{10}{3} \]

### Final Answer:
The value of \( \lambda \) is:
\[ \lambda = -\frac{10}{3} \]