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: Understanding the Concept
The equation $Ax^2+2Hxy+By^2=0$ represents a pair of straight lines passing through the origin. We can find the angle between these lines and the product of their slopes using standard formulas derived from this general equation.
Step 2: Key Formula or Approach
1. For the general equation $Ax^2+2Hxy+By^2=0$, the angle $\theta$ between the lines is given by $\tan\theta = \left|\frac{2\sqrt{H^2-AB}}{A+B}\right|$. 2. The product of the slopes ($m_1m_2$) of the two lines is given by $m_1m_2 = \frac{A}{B}$. 3. We will use the angle formula to find the integer value of $a$. 4. Then we will use the value of $a$ in the product of slopes formula.
Step 3: Detailed Explanation
1. Find the value of a: The given equation is $ax^2+4xy-2y^2=0$. Comparing with $Ax^2+2Hxy+By^2=0$, we have $A=a$, $2H=4 \implies H=2$, and $B=-2$. The angle is $\theta = \text{Tan}^{-1}(2\sqrt{10})$, which means $\tan\theta = 2\sqrt{10}$. Using the angle formula: \[ 2\sqrt{10} = \left|\frac{2\sqrt{2^2 - a(-2)}}{a + (-2)}\right| = \left|\frac{2\sqrt{4+2a}}{a-2}\right| \] Divide by 2: \[ \sqrt{10} = \left|\frac{\sqrt{4+2a}}{a-2}\right| \] Square both sides: \[ 10 = \frac{4+2a}{(a-2)^2} \] \[ 10(a^2 - 4a + 4) = 4+2a \] \[ 10a^2 - 40a + 40 = 4+2a \] \[ 10a^2 - 42a + 36 = 0 \] Divide by 2: \[ 5a^2 - 21a + 18 = 0 \] We solve this quadratic equation for $a$. Since $a \in \mathbb{Z}$, we can test integer factors or use the quadratic formula. \[ a = \frac{-(-21) \pm \sqrt{(-21)^2 - 4(5)(18)}}{2(5)} = \frac{21 \pm \sqrt{441 - 360}}{10} = \frac{21 \pm \sqrt{81}}{10} = \frac{21 \pm 9}{10} \] The two possible values for $a$ are: $a_1 = \frac{21+9}{10} = \frac{30}{10} = 3$. This is an integer. $a_2 = \frac{21-9}{10} = \frac{12}{10} = 1.2$. This is not an integer. Therefore, the only valid value is $a=3$. 2. Find the product of the slopes: The equation of the pair of lines is $3x^2+4xy-2y^2=0$. The product of the slopes is given by the formula $m_1m_2 = \frac{A}{B}$. Here, $A=a=3$ and $B=-2$. \[ m_1m_2 = \frac{3}{-2} = -\frac{3}{2} \] Step 4: Final Answer
The integer value of $a$ is 3, and the product of the slopes of the lines is $-\frac{3}{2}$.