Step 1: Understanding the Concept:
A homogeneous equation of second degree $ax^2 + 2hxy + by^2 = 0$ represents a pair of straight lines passing through the origin. If their slopes are $m_1$ and $m_2$, there exist direct relations linking the sum and product of these slopes to the coefficients $a, h$, and $b$. We substitute these relations into the given condition to find the specific ratio between the slopes.
Step 2: Key Formula or Approach:
1. Sum of slopes: $m_1 + m_2 = -\frac{2h}{b}$.
2. Product of slopes: $m_1 m_2 = \frac{a}{b}$.
3. Express the given condition $16h^2 = 25ab$ entirely in terms of slopes.
Step 3: Detailed Explanation:
From the standard formulas, we can express $h$ and $a$ in terms of the slopes and $b$:
\[ 2h = -b(m_1 + m_2) \implies h = -\frac{b(m_1 + m_2)}{2} \]
\[ a = b(m_1 m_2) \]
The given condition is:
\[ 16h^2 = 25ab \]
Substitute the expressions for $h$ and $a$ into this condition:
\[ 16 \left[ -\frac{b(m_1 + m_2)}{2} \right]^2 = 25 [b(m_1 m_2)] b \]
Square the term in the brackets:
\[ 16 \left[ \frac{b^2(m_1 + m_2)^2}{4} \right] = 25 b^2 (m_1 m_2) \]
Simplify the left side ($16/4 = 4$):
\[ 4 b^2 (m_1 + m_2)^2 = 25 b^2 (m_1 m_2) \]
Assuming $b \neq 0$ (otherwise it's not a proper pair of lines with finite slopes), divide by $b^2$:
\[ 4(m_1 + m_2)^2 = 25(m_1 m_2) \]
Expand the squared binomial:
\[ 4(m_1^2 + 2m_1m_2 + m_2^2) = 25m_1m_2 \]
\[ 4m_1^2 + 8m_1m_2 + 4m_2^2 = 25m_1m_2 \]
Rearrange into a homogeneous quadratic equation in terms of $m_1$ and $m_2$:
\[ 4m_1^2 - 17m_1m_2 + 4m_2^2 = 0 \]
To find the relationship ratio, divide the entire equation by $m_2^2$ (assuming $m_2 \neq 0$):
\[ 4\left(\frac{m_1}{m_2}\right)^2 - 17\left(\frac{m_1}{m_2}\right) + 4 = 0 \]
Let the ratio $r = \frac{m_1}{m_2}$. The equation becomes:
\[ 4r^2 - 17r + 4 = 0 \]
Factor the quadratic equation:
\[ 4r^2 - 16r - r + 4 = 0 \]
\[ 4r(r - 4) - 1(r - 4) = 0 \]
\[ (4r - 1)(r - 4) = 0 \]
So, $r = \frac{1}{4}$ or $r = 4$.
This means $\frac{m_1}{m_2} = \frac{1}{4} \implies 4m_1 = m_2$, OR $\frac{m_1}{m_2} = 4 \implies m_1 = 4m_2$.
Looking at the options, $m_1 = 4m_2$ is present. Since the naming of slopes is arbitrary, both relations denote the same physical pair of lines where one slope is four times the other.
Step 4: Final Answer:
The relation is $m_1 = 4m_2$.
