Question

If the axes are rotated through an angle \( \alpha \), then the number of values of \( \alpha \) such that the transformed equation of \( x^2 + y^2 + 2x + 2y - 5 = 0 \) contains no linear terms is:

• A. \( 0 \)
• B. \( 1 \)
• C. \( 2 \)
• D. Infinite.

Hint:

When rotating a conic section, the linear terms disappear when the sum of their coefficients equals zero. Solve for \( \alpha \) using the tangent identity to find valid rotation angles.

Answer 1

The Correct Option is B

Step 1: Condition for Removal of Linear Terms
When the equation of a conic section is rotated, the linear terms \( x' \) and \( y' \) disappear if the transformed equation does not contain any first-degree terms. This happens when the coefficients of \( x \) and \( y \) are eliminated by choosing an appropriate rotation angle \( \alpha \).
Step 2: General Transformation Equations The given equation is: \[ x^2 + y^2 + 2x + 2y - 5 = 0. \] Under a rotation by \( \alpha \), the new coordinates are: \[ x = x' \cos\alpha - y' \sin\alpha, \quad y = x' \sin\alpha + y' \cos\alpha. \] Substituting these into the given equation and simplifying, we ensure that the linear terms vanish. The linear coefficients in terms of \( \alpha \) are given by: \[ 2\cos\alpha + 2\sin\alpha = 0. \] Step 3: Solving for \( \alpha \) \[ \cos\alpha + \sin\alpha = 0. \] Dividing by \( \cos\alpha \): \[ 1 + \tan\alpha = 0. \] \[ \tan\alpha = -1. \] \[ \alpha = \tan^{-1}(-1) = -\frac{\pi}{4}. \] Since rotation angles are considered in the range \( 0 \leq \alpha<\pi \), the only valid solution is: \[ \alpha = \frac{3\pi}{4}. \] Step 4: Conclusion
Thus, there is exactly one value of \( \alpha \) that satisfies the given condition.
Thus, the correct answer is \( \mathbf{1} \).