Question

If the slopes of the lines represented by the equation $6x^2+2hxy+4y^2 = 0$ are in the ratio 2:3, then the value of h such that both the lines make acute angles with the positive X-axis measured in positive direction is

• A. 5
• B. $5/2$
• C. -5
• D. $-5/2$

Hint:

For a pair of lines $ax^2+2hxy+by^2=0$, remember the formulas for the sum of slopes ($m_1+m_2 = -2h/b$) and the product of slopes ($m_1m_2 = a/b$). If a ratio of slopes is given, let them be $pk$ and $qk$ and solve for $k$.

Answer 1

The Correct Option is C

The given equation is a homogeneous equation of second degree representing a pair of lines through the origin: $6x^2+2hxy+4y^2 = 0$.
Let the slopes of the two lines be $m_1$ and $m_2$.
From the general equation $ax^2+2hxy+by^2=0$, we have:
Sum of slopes: $m_1 + m_2 = -\frac{2h}{b} = -\frac{2h}{4} = -\frac{h}{2}$.
Product of slopes: $m_1 m_2 = \frac{a}{b} = \frac{6}{4} = \frac{3}{2}$.
We are given that the slopes are in the ratio 2:3. Let the slopes be $2k$ and $3k$.
From the product of slopes:
$(2k)(3k) = \frac{3}{2} \implies 6k^2 = \frac{3}{2} \implies k^2 = \frac{3}{12} = \frac{1}{4}$.
So, $k = \pm\frac{1}{2}$.
Now, from the sum of slopes:
$m_1 + m_2 = 2k + 3k = 5k = -\frac{h}{2}$.
This gives $h = -10k$.
Case 1: $k = \frac{1}{2}$. $h = -10(\frac{1}{2}) = -5$. The slopes are $m_1 = 2(\frac{1}{2})=1$ and $m_2 = 3(\frac{1}{2})=\frac{3}{2}$.
Case 2: $k = -\frac{1}{2}$. $h = -10(-\frac{1}{2}) = 5$. The slopes are $m_1 = 2(-\frac{1}{2})=-1$ and $m_2 = 3(-\frac{1}{2})=-\frac{3}{2}$.
The problem states that both lines make acute angles with the positive X-axis. This means both slopes must be positive.
Case 1 gives positive slopes ($1$ and $3/2$), while Case 2 gives negative slopes.
Therefore, we must choose Case 1, which corresponds to $h = -5$.