Question

If $\text{Tan}^{-1}(2\sqrt{10})$ is the angle between the lines $ax^2+4xy-2y^2=0$ and $a \in \mathbb{Z}$, then the product of the slopes of given lines is

• A. $\frac{3}{2}$
• B. $\frac{2}{3}$
• C. $-\frac{2}{3}$
• D. $-\frac{3}{2}$

Hint:

For a homogeneous second-degree equation $Ax^2+2Hxy+By^2=0$, remember the sum of slopes is $m_1+m_2 = -2H/B$ and the product of slopes is $m_1m_2 = A/B$. These are quick to apply once you have all the coefficients.

Answer 1

The Correct Option is D

Step 1: Angle Formula for Pair of Lines: For \( Ax^2+2Hxy+By^2=0 \), the angle \( \theta \) satisfies \( \tan\theta = \frac{2\sqrt{H^2-AB}}{|A+B|} \). Here \( A=a, 2H=4 \implies H=2, B=-2 \). Given \( \tan\theta = 2\sqrt{10} \).
Step 2: Solve for 'a': \[ 2\sqrt{10} = \frac{2\sqrt{2^2 - a(-2)}}{|a-2|} \] \[ \sqrt{10} = \frac{\sqrt{4+2a}}{|a-2|} \] Square both sides: \[ 10 = \frac{4+2a}{(a-2)^2} \] \[ 5 = \frac{2+a}{(a-2)^2} \] \[ 5(a^2-4a+4) = 2+a \] \[ 5a^2 - 20a + 20 = 2 + a \] \[ 5a^2 - 21a + 18 = 0 \] Solving quadratic: \( a = \frac{21 \pm \sqrt{441 - 360}}{10} = \frac{21 \pm 9}{10} \). \( a = 3 \) or \( a = 1.2 \). Since \( a \in \mathbb{Z} \), \( a=3 \).
Step 3: Product of Slopes: For \( Ax^2+2Hxy+By^2=0 \), product of slopes \( m_1m_2 = \frac{A}{B} \). \[ m_1m_2 = \frac{a}{-2} = \frac{3}{-2} = -\frac{3}{2} \]