Question

If angle between the lines represented by \( ax^2 + 2hxy + by^2 = 0 \) is equal to the angle between the lines represented by \( 2x^2 - 5xy + 3y^2 = 0 \), then show that \( 100 (h^2 - ab) = (a + b)^2 \).

Hint:

For pair of lines, use \( \tan \theta = \frac{2 \sqrt{h^2 - ab}}{|a + b|} \); equate angles to relate coefficients.

Answer 1

For pair of lines \( ax^2 + 2hxy + by^2 = 0 \), the angle \( \theta \) between them is: 
\[ \tan \theta = \frac{2 \sqrt{h^2 - ab}}{|a + b|}. \] For \( 2x^2 - 5xy + 3y^2 = 0 \): \( a = 2 \), \( 2h = -5 \Rightarrow h = -\frac{5}{2} \), \( b = 3 \). 
\[ h^2 - ab = \left( -\frac{5}{2} \right)^2 - 2 \cdot 3 = \frac{25}{4} - 6 = \frac{25 - 24}{4} = \frac{1}{4}. \] \[ \sqrt{h^2 - ab} = \sqrt{\frac{1}{4}} = \frac{1}{2}, a + b = 2 + 3 = 5. \] \[ \tan \theta = \frac{2 \cdot \frac{1}{2}}{|5|} = \frac{1}{5}. \] For \( ax^2 + 2hxy + by^2 = 0 \), let the angle be the same: 
\[ \frac{2 \sqrt{h^2 - ab}}{|a + b|} = \frac{1}{5}. \] \[ 2 \sqrt{h^2 - ab} = \frac{|a + b|}{5} \Rightarrow \sqrt{h^2 - ab} = \frac{|a + b|}{10}. \] Square both sides: 
\[ h^2 - ab = \frac{(a + b)^2}{100}. \] \[ 100 (h^2 - ab) = (a + b)^2. \] Answer: Shown.