Question

If the lines $x^2 - 4xy + y^2 = 0$ make angles $\alpha$ and $\beta$ with the positive direction of the $X$-axis, then $\cot^2\alpha + \cot^2\beta =$

• A. 14
• B. 16
• C. 18
• D. 20

Hint:

Whenever the product of slopes is $m_1 \cdot m_2 = 1$, it means the two lines are symmetric reflections across the line $y = x$. Because of this reciprocal symmetry, any symmetric expression in cotangent will yield the exact same value in tangent: $\cot^2\alpha + \cot^2\beta = \tan^2\alpha + \tan^2\beta = \left(\frac{2h}{b}\right)^2 - 2\left(\frac{a}{b}\right)$. Here, $4^2 - 2 = 14$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem presents a homogeneous second-degree equation representing a pair of straight lines passing through the origin. These lines make angles $\alpha$ and $\beta$ with the positive $X$-axis. We need to find the value of the trigonometric expression $\cot^2\alpha + \cot^2\beta$.

Step 2: Key Formula or Approach:
The slopes of the lines are $m_1 = \tan\alpha$ and $m_2 = \tan\beta$. For a standard pair of straight lines equation $ax^2 + 2hxy + by^2 = 0$, the relations for the sum and product of slopes are:
$$m_1 + m_2 = -\frac{2h}{b}, \quad m_1 \cdot m_2 = \frac{a}{b}$$ Here, comparing coefficients: $a = 1, 2h = -4, b = 1$.
We will convert our target expression from cotangent to tangent terms to use these slope relationships.

Step 3: Detailed Explanation:
First, calculate the sum and product of the slopes:
$$\tan\alpha + \tan\beta = -\frac{-4}{1} = 4$$ $$\tan\alpha \cdot \tan\beta = \frac{1}{1} = 1$$ Since $\tan\alpha \cdot \tan\beta = 1$, we can deduce that $\cot\alpha = \tan\beta$ and $\cot\beta = \tan\alpha$.
Let's rewrite our target expression using this substitution:
$$\cot^2\alpha + \cot^2\beta = \text{tan}^2\beta + \text{tan}^2\alpha = \tan^2\alpha + \tan^2\beta$$ Using the algebraic identity $x^2 + y^2 = (x + y)^2 - 2xy$, we expand the tangent terms:
$$\tan^2\alpha + \tan^2\beta = (\tan\alpha + \tan\beta)^2 - 2\tan\alpha\tan\beta$$ Substitute our sum and product slope values into this equation:
$$\tan^2\alpha + \tan^2\beta = (4)^2 - 2(1) = 16 - 2 = 14$$ Therefore, $\cot^2\alpha + \cot^2\beta = 14$.

Step 4: Final Answer:
The value of the expression is 14, which corresponds to option (A).