When the axes of coordinates are rotates through an angle `pi//4` without shifting the origin, the equation `2x^(2)+2y^(2)+3xy=4` is transformed to the equation `7x^(2)+y^(2)=k` where the value of k is

Answer

To solve the problem, we need to transform the equation \(2x^2 + 2y^2 + 3xy = 4\) when the coordinate axes are rotated through an angle of \(\frac{\pi}{4}\). The transformed equation is given as \(7x^2 + y^2 = k\), and we need to find the value of \(k\).

### Step-by-Step Solution:

1. **Rotation of Axes Transformation**:
When the axes are rotated by an angle \(\theta\), the new coordinates \((x', y')\) can be expressed in terms of the old coordinates \((x, y)\) as follows:
\[ x' = x \cos \theta + y \sin \theta \]
\[ y' = -x \sin \theta + y \cos \theta \]
For \(\theta = \frac{\pi}{4}\), we have:
\[ \cos \frac{\pi}{4} = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \]
Thus, the transformations become:
\[ x' = \frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = \frac{x + y}{\sqrt{2}} \]
\[ y' = -\frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = \frac{y - x}{\sqrt{2}} \]

2. **Substituting into the Original Equation**:
We substitute \(x\) and \(y\) in terms of \(x'\) and \(y'\) into the original equation \(2x^2 + 2y^2 + 3xy = 4\). First, we express \(x\) and \(y\) in terms of \(x'\) and \(y'\):
\[ x = \frac{x' - y'}{\sqrt{2}}, \quad y = \frac{x' + y'}{\sqrt{2}} \]

3. **Expanding the Original Equation**:
Substitute \(x\) and \(y\) into the original equation:
\[ 2\left(\frac{x' - y'}{\sqrt{2}}\right)^2 + 2\left(\frac{x' + y'}{\sqrt{2}}\right)^2 + 3\left(\frac{x' - y'}{\sqrt{2}}\right)\left(\frac{x' + y'}{\sqrt{2}}\right) = 4 \]
Expanding each term:
\[ 2\left(\frac{x'^2 - 2x'y' + y'^2}{2}\right) + 2\left(\frac{x'^2 + 2x'y' + y'^2}{2}\right) + 3\left(\frac{x'^2 - y'^2}{2}\right) = 4 \]
Simplifying:
\[ (x'^2 - 2x'y' + y'^2) + (x'^2 + 2x'y' + y'^2) + \frac{3}{2}(x'^2 - y'^2) = 4 \]
Combining like terms:
\[ 2x'^2 + 2y'^2 + \frac{3}{2}x'^2 - \frac{3}{2}y'^2 = 4 \]
This simplifies to:
\[ \left(2 + \frac{3}{2}\right)x'^2 + \left(2 - \frac{3}{2}\right)y'^2 = 4 \]
\[ \frac{7}{2}x'^2 + \frac{1}{2}y'^2 = 4 \]

4. **Multiplying through by 2**:
To eliminate the fractions, multiply the entire equation by 2:
\[ 7x'^2 + y'^2 = 8 \]

5. **Finding \(k\)**:
We compare this with the equation \(7x^2 + y^2 = k\). From our transformation, we see that \(k = 8\).

### Final Answer:
The value of \(k\) is \(8\).