Question

If the lines represented by $ax^2 - bxy - y^2 = 0$ make angles $\alpha$ and $\beta$ with the positive direction of the $X$-axis, then $\tan(\alpha + \beta) = $

• A. $\frac{a}{a+b}$
• B. $\frac{b}{1+b}$
• C. $\frac{b}{1+a}$
• D. $-\frac{b}{1+a}$

Hint:

When evaluating trigonometric slope properties for pair of straight lines, always derive $m_1+m_2$ and $m_1m_2$ directly from coefficients. Be exceptionally careful with negative signs in the denominator when $B = -1$.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The given equation $ax^2 - bxy - y^2 = 0$ represents a pair of straight lines passing through the origin. These lines make angles $\alpha$ and $\beta$ with the positive $X$-axis, meaning their slopes are $m_1 = \tan\alpha$ and $m_2 = \tan\beta$. We need to compute the value of $\tan(\alpha + \beta)$.

Step 2: Key Formula or Approach:
For a standard homogeneous second-degree equation of a pair of lines $Ax^2 + 2Hxy + By^2 = 0$, the sum and product of the slopes $m_1$ and $m_2$ are given by: $$m_1 + m_2 = -\frac{2H}{B} \quad \text{and} \quad m_1m_2 = \frac{A}{B}$$ The compound angle formula for tangent is: $$\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} = \frac{m_1 + m_2}{1 - m_1m_2}$$

Step 3: Detailed Explanation:
Comparing the given equation $ax^2 - bxy - y^2 = 0$ with the standard form $Ax^2 + 2Hxy + By^2 = 0$, we get: $$A = a, \quad 2H = -b, \quad B = -1$$ Now, compute the sum and product of the slopes: $$m_1 + m_2 = -\frac{-b}{-1} = -b$$ $$m_1m_2 = \frac{a}{-1} = -a$$ Substitute these slope relations into the compound angle expansion formula: $$\tan(\alpha + \beta) = \frac{-b}{1 - (-a)} = -\frac{b}{1+a}$$ This matches the expression provided in option (D).

Step 4: Final Answer:
The value of $\tan(\alpha + \beta)$ is $-\frac{b}{1+a}$, which corresponds to option (D).