Find the point at which origin is shifted such that the transformed equation of `x^(2)+2y^(2)-4x+4y-2=0` has no first degree term. Also find the transformed equation .

Answer

To find the point at which the origin is shifted such that the transformed equation of \(x^2 + 2y^2 - 4x + 4y - 2 = 0\) has no first degree term, we will follow these steps:

### Step 1: Understand the Shift of Origin
When we shift the origin from \((0, 0)\) to a new point \((h, k)\), the new coordinates \((X, Y)\) in terms of the old coordinates \((x, y)\) are given by:
\[ X = x - h \quad \text{and} \quad Y = y - k \]

### Step 2: Substitute the New Coordinates
We will substitute \(x\) and \(y\) in terms of \(X\) and \(Y\) into the given equation:
\[ x = X + h \quad \text{and} \quad y = Y + k \]
Substituting these into the original equation:
\[ (X + h)^2 + 2(Y + k)^2 - 4(X + h) + 4(Y + k) - 2 = 0 \]

### Step 3: Expand the Equation
Now we will expand the equation:
\[ (X^2 + 2hX + h^2) + 2(Y^2 + 2kY + k^2) - 4X - 4h + 4Y + 4k - 2 = 0 \]
This simplifies to:
\[ X^2 + 2Y^2 + (2h - 4)X + (4 + 4k)Y + (h^2 + 2k^2 - 4h + 4k - 2) = 0 \]

### Step 4: Set Coefficients of First Degree Terms to Zero
For the transformed equation to have no first degree terms, the coefficients of \(X\) and \(Y\) must be zero:
1. \(2h - 4 = 0\)
2. \(4 + 4k = 0\)

### Step 5: Solve for \(h\) and \(k\)
From the first equation:
\[ 2h - 4 = 0 \implies h = 2 \]
From the second equation:
\[ 4 + 4k = 0 \implies 4k = -4 \implies k = -1 \]

### Step 6: Find the Transformed Equation
Now substituting \(h = 2\) and \(k = -1\) back into the expanded equation:
\[ X^2 + 2Y^2 + (0)X + (0)Y + (2^2 + 2(-1)^2 - 4(2) + 4(-1) - 2) = 0 \]
Calculating the constant term:
\[ 2^2 + 2(1) - 8 - 4 - 2 = 4 + 2 - 8 - 4 - 2 = -8 \]
Thus, the transformed equation is:
\[ X^2 + 2Y^2 - 8 = 0 \]

### Final Answer
The point at which the origin is shifted is \((2, -1)\) and the transformed equation is:
\[ X^2 + 2Y^2 - 8 = 0 \]