Step 1: Understanding the Concept:
To eliminate the first-degree terms (linear terms in \( x \) and \( y \)), the origin must be shifted to the center \( (h,k) \) of the conic. The center is obtained by equating the partial derivatives of the curve's equation to zero.
To eliminate the \( xy \) term, the axes must be rotated by an angle \( \theta \) such that \( \tan 2\theta = \frac{2H}{A-B} \) for the general equation \( Ax^2 + 2Hxy + By^2 + \dots = 0 \).
Step 2: Key Formula or Approach:
1. Center \( (h,k) \): Solve \( \frac{\partial S}{\partial x} = 0 \) and \( \frac{\partial S}{\partial y} = 0 \).
2. Angle of rotation: \( \tan 2\theta = \frac{2H}{A-B} \).
Step 3: Detailed Explanation:
Given \( S \equiv 2x^2 - xy - y^2 - 3x + 3y = 0 \).
Comparing with \( Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0 \):
\( A = 2 \), \( 2H = -1 \implies H = -1/2 \), \( B = -1 \).
Calculate Center (h, k):
Partial derivative w.r.t \( x \):
\( 4x - y - 3 = 0 \implies 4x - y = 3 \) ... (i)
Partial derivative w.r.t \( y \):
\( -x - 2y + 3 = 0 \implies x + 2y = 3 \) ... (ii)
Multiply (i) by 2: \( 8x - 2y = 6 \). Add to (ii):
\( 9x = 9 \implies x = 1 \).
Substitute \( x = 1 \) in (ii):
\( 1 + 2y = 3 \implies 2y = 2 \implies y = 1 \).
So, \( h = 1 \) and \( k = 1 \).
Calculate \( \tan 2\theta \):
\[ \tan 2\theta = \frac{2H}{A - B} = \frac{-1}{2 - (-1)} = \frac{-1}{3} \]
Check Options:
We need an expression equal to \( -1/3 \) using \( h=1, k=1 \).
(A) \( h+k = 2 \)
(B) \( h-k = 0 \)
(C) \( hk = 1 \)
(D) \( -\frac{h}{3k} = -\frac{1}{3(1)} = -\frac{1}{3} \)
Option (D) matches.
Step 4: Final Answer:
The value is \( -\frac{h}{3k} \).