Question

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

• A. $\frac{2h}{b - a}$
• B. $\frac{2h}{a - b}$
• C. $\frac{h}{a + b}$
• D. $\frac{2h}{a + b}$

Hint:

To remember this formula easily, notice that the lines are perpendicular when $a + b = 0 \implies a = -b$. If $a = -b$, the denominator of our answer becomes $-b - b = -2b$, making the whole tangent expression reduce to $-\frac{h}{b}$. Keeping trace configurations in mind helps verify algebraic structures quickly!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The equation $ax^2 + 2hxy + by^2 = 0$ represents a pair of straight lines passing through the origin. We are given that these two lines have angles of inclination $\alpha$ and $\beta$ with the positive $x$-axis, and we need to find the expression for $\tan(\alpha + \beta)$.

Step 2: Key Formula or Approach:
1. The slopes of the lines are $m_1 = \tan \alpha$ and $m_2 = \tan \beta$.
2. For a homogeneous second-degree equation, the sum and product of the slopes are governed by the relations: $$m_1 + m_2 = -\frac{2h}{b} \quad \text{and} \quad m_1 \cdot m_2 = \frac{a}{b}$$ 3. Use the standard trigonometric compound angle identity: $$\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \cdot \tan \beta} = \frac{m_1 + m_2}{1 - m_1 m_2}$$

Step 3: Detailed Explanation:
Substitute the slope formulas directly into the trigonometric identity: $$\tan(\alpha + \beta) = \frac{-\frac{2h}{b}}{1 - \frac{a}{b}}$$ Simplify the denominator fraction by creating a common base: $$1 - \frac{a}{b} = \frac{b - a}{b}$$ Now divide the two fractional expressions: $$\tan(\alpha + \beta) = \frac{-\frac{2h}{b}}{\frac{b - a}{b}} = \frac{-2h}{b - a}$$ Distribute the negative sign in the denominator to flip the subtraction terms: $$\tan(\alpha + \beta) = \frac{2h}{a - b}$$

Step 4: Final Answer:
The value of $\tan(\alpha + \beta)$ is $\frac{2h}{a - b}$, which matches option (B).