Question

The slopes of the lines represented by $6x^2 + 2hxy + y^2 = 0$ are in the ratio 2 : 3, then $h = \dots$

• A. $\pm 7/2$
• B. $\pm 1/2$
• C. $\pm 5/2$
• D. $\pm 2/5$

Hint:

There is a direct shortcut formula! If the slopes of $ax^2 + 2hxy + by^2 = 0$ are in the ratio $m:n$, then:
$\frac{(2h)^2}{ab} = \frac{(m+n)^2}{mn}$.
Here: $\frac{4h^2}{(6)(1)} = \frac{(2+3)^2}{(2)(3)} \implies \frac{4h^2}{6} = \frac{25}{6} \implies 4h^2 = 25 \implies h = \pm 5/2$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given a homogeneous equation of degree 2 representing a pair of straight lines passing through the origin. The ratio of their slopes is 2:3. We must find the value of the unknown coefficient $h$.

Step 2: Key Formula or Approach:
For the standard equation $ax^2 + 2hxy + by^2 = 0$, if the individual lines have slopes $m_1$ and $m_2$, they satisfy two relationships:
1. Sum of slopes: $m_1 + m_2 = \frac{-2h}{b}$
2. Product of slopes: $m_1 m_2 = \frac{a}{b}$

Step 3: Detailed Explanation:
From the given equation $6x^2 + 2hxy + 1y^2 = 0$, we identify $a = 6$, $2h = 2h$, and $b = 1$.
Since the slopes are in a 2:3 ratio, let's denote them as $m_1 = 2k$ and $m_2 = 3k$, where $k$ is an arbitrary constant multiplier.
Apply the product rule:
$$m_1 \cdot m_2 = \frac{a}{b} = \frac{6}{1} = 6$$
$$(2k)(3k) = 6 \implies 6k^2 = 6 \implies k^2 = 1 \implies k = \pm 1$$
Now apply the sum rule:
$$m_1 + m_2 = \frac{-2h}{b} = \frac{-2h}{1} = -2h$$
$$(2k) + (3k) = -2h \implies 5k = -2h$$
Isolate $h$:
$$h = -\frac{5k}{2}$$
Substitute the values we found for $k = \pm 1$:
If $k = 1 \implies h = -5/2$.
If $k = -1 \implies h = 5/2$.
Thus, the possible values for $h$ are $\pm 5/2$.

Step 4: Final Answer:
The value of $h$ is $\pm 5/2$, matching option (c).