Find the equation of the curve `2x^(2)+y^(2)-3x+5y-8=0` when the origin is transferred to the point `(-1, 2)` without changing the direction of axes.

Answer

To find the equation of the curve \( 2x^2 + y^2 - 3x + 5y - 8 = 0 \) when the origin is transferred to the point \((-1, 2)\) without changing the direction of the axes, we will follow these steps:

### Step 1: Understand the Transformation
When the origin is shifted to the point \((-1, 2)\), we need to express the new coordinates \( (x, y) \) in terms of the old coordinates \( (X, Y) \). The transformations are given by:
\[ x = X + 1 \quad \text{and} \quad y = Y - 2 \]

### Step 2: Substitute the Transformations into the Original Equation
We will substitute \( x = X + 1 \) and \( y = Y - 2 \) into the original equation:
\[ 2(X + 1)^2 + (Y - 2)^2 - 3(X + 1) + 5(Y - 2) - 8 = 0 \]

### Step 3: Expand the Equation
Now we will expand each term:
1. \( (X + 1)^2 = X^2 + 2X + 1 \)
\[ 2(X + 1)^2 = 2(X^2 + 2X + 1) = 2X^2 + 4X + 2 \]
2. \( (Y - 2)^2 = Y^2 - 4Y + 4 \)
3. \( -3(X + 1) = -3X - 3 \)
4. \( 5(Y - 2) = 5Y - 10 \)

Now substituting these back into the equation:
\[ 2X^2 + 4X + 2 + Y^2 - 4Y + 4 - 3X - 3 + 5Y - 10 - 8 = 0 \]

### Step 4: Combine Like Terms
Now we combine all the like terms:
- For \( X^2 \): \( 2X^2 \)
- For \( X \): \( 4X - 3X = 1X \)
- For \( Y^2 \): \( Y^2 \)
- For \( Y \): \( -4Y + 5Y = 1Y \)
- Constant terms: \( 2 + 4 - 3 - 10 - 8 = -15 \)

Putting it all together, we have:
\[ 2X^2 + X + Y^2 + Y - 15 = 0 \]

### Step 5: Replace \( X \) and \( Y \) with \( x \) and \( y \)
Finally, we replace \( X \) with \( x \) and \( Y \) with \( y \):
\[ 2x^2 + x + y^2 + y - 15 = 0 \]

### Final Answer
The equation of the curve after the transformation is:
\[ 2x^2 + x + y^2 + y - 15 = 0 \]