Question

By rotating the axes about the origin in anti-clockwise direction with certain angle, if the equation \( x^2 + 4xy + y^2 = 1 \) is transformed to \( \frac{x'^2}{a^2} - \frac{y'^2}{b^2} = 1 \), then \( \sqrt{\frac{a^2 + b^2}{a^2}} = \)

• A. \( 2 \)
• B. \( \frac{\sqrt{13}}{3} \)
• C. \( \frac{3}{2} \)
• D. \( \sqrt{10} \)

Hint:

When a conic is rotated to eliminate the \( xy \)-term, its classification and axis lengths can be determined using eigenvalues of its coefficient matrix.

Answer 1

The Correct Option is A

Step 1: General second-degree equation form 
The given equation is: 
\[ x^2 + 4xy + y^2 = 1 \] This is a conic in the general second-degree form: 
\[ Ax^2 + 2Hxy + By^2 = 1 \] where \( A = 1 \), \( 2H = 4 \Rightarrow H = 2 \), and \( B = 1 \) 
Step 2: Use rotation of axes to eliminate the \( xy \)-term 
Under a rotation of axes by angle \( \theta \), the new coefficients \( A', B' \) are the eigenvalues of the matrix: 
\[ \begin{bmatrix} A & H \\ H & B \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} \] Step 3: Find eigenvalues 
Solve the characteristic equation: 
\[ \begin{vmatrix} 1 - \lambda & 2 \\ 2 & 1 - \lambda \end{vmatrix} = (1 - \lambda)^2 - 4 = \lambda^2 - 2\lambda - 3 = 0 \] \[ \Rightarrow \lambda = 3, -1 \] Step 4: Use transformed conic form 
The equation becomes: 
\[ \frac{x'^2}{a^2} - \frac{y'^2}{b^2} = 1 \Rightarrow \text{a hyperbola, so eigenvalues are } \frac{1}{a^2} = 3, \quad \frac{1}{b^2} = 1 \] Step 5: Solve for \( a^2 \) and \( b^2 \) 
\[ a^2 = \frac{1}{3}, \quad b^2 = 1 \Rightarrow a^2 + b^2 = \frac{1}{3} + 1 = \frac{4}{3} \] Step 6: Find the expression 
\[ \sqrt{\frac{a^2 + b^2}{a^2}} = \sqrt{ \frac{\frac{4}{3}}{\frac{1}{3}} } = \sqrt{4} = 2 \] 
Hence, 
\[ \boxed{2} \]