Question

Suppose the axes are to be rotated through an angle \( \theta \) so as to remove the \( xy \) term from the equation \(3 x^2 + 2\sqrt{3}xy + y^2 = 0 \). Then in the new coordinate system, the equation \( x^2 + y^2 + 2xy = 2 \) is transformed to:

• A. \( (2 + \sqrt{3})x^2 + (2 - \sqrt{3})y^2 + 2xy = 4 \)
• B. \( (2 - \sqrt{3})x^2 + (2 + \sqrt{3})y^2 - 2xy = 4 \)
• C. \( x^2 + y^2 - 2(2 - \sqrt{3})xy = 4(2 - \sqrt{3}) \)
• D. \( x^2 + y^2 + 2(2 + \sqrt{3})xy = 4(2 + \sqrt{3}) \)

Hint:

To remove the \( xy \)-term in a conic equation, use the rotation of axes technique with the angle \( \theta \) such that \( \tan(2\theta) = \frac{2B}{A - C} \), where \( A \), \( B \), and \( C \) are the coefficients of the quadratic terms.

Answer 1

The Correct Option is A

Step 1: Rotate the coordinate system. We are given the equation \( 3x^2 + 2\sqrt{3}xy + y^2 = 0 \) and we need to remove the \( xy \)-term by rotating the coordinate system. The angle \( \theta \) of rotation is given by: \[ \tan 2\theta = \frac{2B}{A - C} \] where \( A = 3 \), \( B = \sqrt{3} \), and \( C = 1 \). Substituting these values: \[ \tan 2\theta = \frac{2\sqrt{3}}{2} = \sqrt{3}. \] Thus, \( \theta = 45^\circ \). Step 2: Apply the transformation. Using the formulas for coordinate rotation, we find the transformed equation: \[ (2 + \sqrt{3})x^2 + (2 - \sqrt{3})y^2 + 2xy = 4. \]

Answer 2

Given the equation:
\[ 3 x^2 + 2 \sqrt{3} x y + y^2 = 0 \] The axes are rotated by angle \( \theta \) to remove the \( xy \) term.

Step 1: The angle \( \theta \) of rotation to remove the \( xy \) term in a conic:
\[ \tan 2\theta = \frac{2 B}{A - C} \] Here, \( A = 3 \), \( B = \sqrt{3} \), and \( C = 1 \). Substitute:
\[ \tan 2\theta = \frac{2 \times \sqrt{3}}{3 - 1} = \frac{2 \sqrt{3}}{2} = \sqrt{3} \] So:
\[ 2\theta = 60^\circ \implies \theta = 30^\circ \]

Step 2: The second equation:
\[ x^2 + y^2 + 2 x y = 2 \] Rewrite in matrix form:
\[ \begin{bmatrix} x & y \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = 2 \] because \( x^2 + y^2 + 2 x y = (x + y)^2 \).

Step 3: Apply rotation by \( \theta = 30^\circ \):
New coordinates \( (X, Y) \) related to old \( (x, y) \) by:
\[ \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} X \\ Y \end{bmatrix} \] Substitute \( \cos 30^\circ = \frac{\sqrt{3}}{2} \), \( \sin 30^\circ = \frac{1}{2} \):
\[ x = \frac{\sqrt{3}}{2} X - \frac{1}{2} Y, \quad y = \frac{1}{2} X + \frac{\sqrt{3}}{2} Y \]

Step 4: Substitute into the second equation:
\[ (x + y)^2 = 2 \] Calculate \( x + y \):
\[ x + y = \left(\frac{\sqrt{3}}{2} X - \frac{1}{2} Y\right) + \left(\frac{1}{2} X + \frac{\sqrt{3}}{2} Y\right) = \frac{\sqrt{3} + 1}{2} X + \frac{\sqrt{3} - 1}{2} Y \] Square both sides:
\[ \left( \frac{\sqrt{3} + 1}{2} X + \frac{\sqrt{3} - 1}{2} Y \right)^2 = 2 \] Multiply both sides by 4:
\[ \left( (\sqrt{3} + 1) X + (\sqrt{3} - 1) Y \right)^2 = 8 \] Expand:
\[ (\sqrt{3} + 1)^2 X^2 + 2 (\sqrt{3} + 1)(\sqrt{3} - 1) X Y + (\sqrt{3} - 1)^2 Y^2 = 8 \] Calculate squares and product:
\[ (\sqrt{3} + 1)^2 = 3 + 2 \sqrt{3} + 1 = 4 + 2 \sqrt{3} \] \[ (\sqrt{3} - 1)^2 = 3 - 2 \sqrt{3} + 1 = 4 - 2 \sqrt{3} \] \[ (\sqrt{3} + 1)(\sqrt{3} - 1) = 3 - 1 = 2 \] So:
\[ (4 + 2 \sqrt{3}) X^2 + 4 X Y + (4 - 2 \sqrt{3}) Y^2 = 8 \] Divide entire equation by 2:
\[ (2 + \sqrt{3}) X^2 + 2 X Y + (2 - \sqrt{3}) Y^2 = 4 \] Rename \( X, Y \) as \( x, y \) in new coordinate system:
\[ \boxed{ (2 + \sqrt{3}) x^2 + (2 - \sqrt{3}) y^2 + 2 x y = 4 } \]