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: Shift the Origin
Let the new coordinates after shifting the origin be \(X\) and \(Y\). We can express the original coordinates \(x\) and \(y\) in terms of \(X\) and \(Y\) as follows:
\[ x = X + h \quad \text{and} \quad y = Y + k \]
where \(h\) and \(k\) are the shifts in the x and y directions, respectively.

### Step 2: Substitute into the Original Equation
Substituting \(x\) and \(y\) into the original equation:
\[ (X + h)^2 + 2(Y + k)^2 - 4(X + h) + 4(Y + k) - 2 = 0 \]

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

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

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

### Step 6: Determine the New Origin
The new origin is shifted to the point \((h, k) = (2, -1)\).

### Step 7: Find the Transformed Equation
Substituting \(h\) and \(k\) back into the expanded equation:
The constant term becomes:
\[ h^2 + 2k^2 - 4h + 4k - 2 = 2^2 + 2(-1)^2 - 4(2) + 4(-1) - 2 \]
Calculating this:
\[ = 4 + 2 - 8 - 4 - 2 = -8 \]
Thus, the transformed equation is:
\[ X^2 + 2Y^2 - 8 = 0 \]

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