To determine to which point the origin should be shifted in order to eliminate the first-degree terms in the equation \(2x^2 - 3y^2 - 12x + 18y - 4 = 0\), follow these steps:
1. Identify the first-degree terms: The terms involving \(x\) and \(y\) are \(-12x\) and \(18y\).
2. Complete the square for the \(x\)-terms:
[tex]\[
2x^2 - 12x \text{ is part of the expression. Factor out the 2: } 2(x^2 - 6x)
\][/tex]
To complete the square:
[tex]\[
x^2 - 6x \text{ can be written as } (x-3)^2 - 9:
\][/tex]
Hence, we have:
[tex]\[
2(x^2 - 6x) = 2((x-3)^2 - 9) = 2(x-3)^2 - 18
\][/tex]
3. Complete the square for the \(y\)-terms:
[tex]\[
-3y^2 + 18y \text{ is part of the expression. Factor out the -3: } -3(y^2 - 6y)
\][/tex]
To complete the square:
[tex]\[
y^2 - 6y \text{ can be written as } (y-3)^2 - 9:
\][/tex]
Hence, we have:
[tex]\[
-3(y^2 - 6y) = -3((y-3)^2 - 9) = -3(y-3)^2 + 27
\][/tex]
4. Substitute the completed squares back into the original equation:
[tex]\[
2(x - 3)^2 - 18 - 3(y - 3)^2 + 27 - 4 = 0
\][/tex]
5. Simplify the equation:
[tex]\[
2(x - 3)^2 - 3(y - 3)^2 + 5 = 0
\][/tex]
Given this form, the first-degree terms have been successfully eliminated, and the equation is now centered around the point \( (3, 3) \).
Thus, the origin should be shifted to [tex]\(\boxed{(3, 3)}\)[/tex].
