Find the point so that the equation :
`12x^(2)-10xy+2y^(2)+11x-5y+2=0`
referred to parallel axes through it may transform into one from which terms of the first degree in `x` and `y` are absent.

Answer

To find the point such that the given equation \( 12x^2 - 10xy + 2y^2 + 11x - 5y + 2 = 0 \) can be transformed into a form where the first-degree terms in \( x \) and \( y \) are absent, we will follow these steps:

### Step 1: Shift the Origin
Let the new coordinates be defined as:
\[ x = X + h \quad \text{and} \quad y = Y + k \]
where \( (h, k) \) is the point we need to find.

### Step 2: Substitute in the Equation
Substituting \( x \) and \( y \) into the original equation gives:
\[ 12(X + h)^2 - 10(X + h)(Y + k) + 2(Y + k)^2 + 11(X + h) - 5(Y + k) + 2 = 0 \]

### Step 3: Expand the Equation
Now, we will expand each term:
1. \( 12(X + h)^2 = 12(X^2 + 2hX + h^2) = 12X^2 + 24hX + 12h^2 \)
2. \( -10(X + h)(Y + k) = -10(XY + kX + hY + hk) = -10XY - 10kX - 10hY - 10hk \)
3. \( 2(Y + k)^2 = 2(Y^2 + 2kY + k^2) = 2Y^2 + 4kY + 2k^2 \)
4. \( 11(X + h) = 11X + 11h \)
5. \( -5(Y + k) = -5Y - 5k \)

Combining all these, we have:
\[ 12X^2 + 24hX + 12h^2 - 10XY - 10kX - 10hY - 10hk + 2Y^2 + 4kY + 2k^2 + 11X + 11h - 5Y - 5k + 2 = 0 \]

### Step 4: Collect Like Terms
Now, we collect the coefficients of \( X \), \( Y \), and the constant terms:
- Coefficient of \( X \): \( 12X^2 - 10Y + (24h - 10k + 11)X \)
- Coefficient of \( Y \): \( 2Y^2 + (-10h + 4k - 5)Y \)
- Constant term: \( 12h^2 - 10hk + 2k^2 + 11h - 5k + 2 = 0 \)

### Step 5: Set Coefficients of \( X \) and \( Y \) to Zero
To eliminate the first-degree terms, we set the coefficients of \( X \) and \( Y \) to zero:
1. \( 24h - 10k + 11 = 0 \) (Equation 1)
2. \( -10h + 4k - 5 = 0 \) (Equation 2)

### Step 6: Solve the System of Equations
From Equation 2, we can express \( k \) in terms of \( h \):
\[ 4k = 10h + 5 \implies k = \frac{10h + 5}{4} \]

Substituting \( k \) into Equation 1:
\[ 24h - 10\left(\frac{10h + 5}{4}\right) + 11 = 0 \]
Multiplying through by 4 to eliminate the fraction:
\[ 96h - 100h - 50 + 44 = 0 \implies -4h - 6 = 0 \implies h = -\frac{3}{2} \]

Now substituting \( h \) back to find \( k \):
\[ k = \frac{10(-\frac{3}{2}) + 5}{4} = \frac{-15 + 5}{4} = \frac{-10}{4} = -\frac{5}{2} \]

### Final Result
Thus, the point \( (h, k) \) is:
\[ \left(-\frac{3}{2}, -\frac{5}{2}\right) \]