Question

If the origin is shifted to remove the first degree terms from the equation \(2x^2 - 3y^2 + 4xy + 4x + 4y - 14 = 0\), then with respect to this new co-ordinate system, the transformed equation of \(x^2 + y^2 - 3xy + 4y + 3 = 0\) is

• A. \(x^2 + y^2 - 3xy - 2x + y + 6 = 0\)
• B. \(x^2 + y^2 - 3xy - 2x + 7y + 3 = 0\)
• C. \(x^2 + y^2 - 3xy - 2x + y + 4 = 0\)
• D. \(x^2 + y^2 - 3xy - 2x + 7y + 4 = 0\)

Hint:

When transforming equations by shifting the origin, focus on eliminating the linear terms by completing the square. After shifting, substitute the new coordinates into the original equation to get the transformed equation in the new coordinate system.

Answer 1

The Correct Option is D

Step 1: Start by identifying the transformation that eliminates the first degree terms in the equation \(2x^2 - 3y^2 + 4xy + 4x + 4y - 14 = 0\). - The first-degree terms are \(4x + 4y\). To eliminate these terms, we need to shift the origin to a new point \((x_0, y_0)\) where the first degree terms vanish. Step 2: To remove the linear terms \(4x + 4y\), we complete the square for \(x\) and \(y\) to find the coordinates of the new origin \((x_0, y_0)\). The shift required for \(x\) and \(y\) will be: \[ x_0 = -\frac{4}{2(2)} = -1, \quad y_0 = -\frac{4}{2(3)} = -\frac{2}{3}. \] Thus, the new coordinates are \(x = x' + 1\) and \(y = y' + \frac{2}{3}\). 
Step 3: Substitute the new coordinates into the second equation \(x^2 + y^2 - 3xy + 4y + 3 = 0\), where \(x' = x + 1\) and \(y' = y + \frac{2}{3}\). Expanding and simplifying this equation will give the transformed equation in the new coordinate system. 
Step 4: After substituting and simplifying the terms, we obtain the transformed equation: \[ x^2 + y^2 - 3xy - 2x + 7y + 4 = 0. \]