Question

The equation \( x^2 - 3xy + \lambda y^2 + 3x - 5y + 2 = 0 \), where \( \lambda \) is a real number represents a pair of lines. If \( \theta \) is the acute angle between the lines, then \( \frac{\text{cosec}^2 \theta}{\sqrt{10}} = \)}

• A. 10
• B. \( \frac{1}{\sqrt{10}} \)
• C. 2
• D. \( \sqrt{10} \)

Hint:

The angle between lines depends only on the second-degree terms ($ax^2 + 2hxy + by^2$).

Answer 1

The Correct Option is D

Step 1: Find \(\lambda\)
For a pair of lines, $\Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.
Comparing: $a=1, h=-3/2, b=\lambda, g=3/2, f=-5/2, c=2$.
Solving $\Delta=0$ gives $\lambda = 2$.
Step 2: Find \(\tan \theta\)
$\tan \theta = \frac{2\sqrt{h^2 - ab}}{a+b} = \frac{2\sqrt{(-3/2)^2 - (1)(2)}}{1+2} = \frac{2\sqrt{9/4 - 2}}{3} = \frac{2(1/2)}{3} = \frac{1}{3}$.
Step 3: Find \(\text{cosec}^2 \theta\)
If $\tan \theta = 1/3$, then $\cot \theta = 3$.
$\text{cosec}^2 \theta = 1 + \cot^2 \theta = 1 + 9 = 10$.
Step 4: Calculation
Value $= \frac{10}{\sqrt{10}} = \sqrt{10}$.
Final Answer:(D)