Step 1: Understanding the Question:
The question gives a pair of straight lines represented by the homogenous quadratic equation $ax^2 + 2hxy + by^2 = 0$. Given that the ratio of their slopes is $5 : 3$, we need to find the numerical ratio of $h^2 : ab$.
Step 2: Key Formula or Approach:
For a standard homogenous second-degree pair of straight lines, the sum and product of the individual slopes $m_1$ and $m_2$ are given by the relationships:
$$m_1 + m_2 = -\frac{2h}{b} \quad \text{and} \quad m_1m_2 = \frac{a}{b}$$
We are given that $\frac{m_1}{m_2} = \frac{5}{3}$, which allows us to express one slope in terms of the other and substitute it back into the equations.
Step 3: Detailed Explanation:
Let the two slopes be $m_1 = 5k$ and $m_2 = 3k$ for some scalar constant $k$.
1. Find the sum of the slopes in terms of $k$:
$$m_1 + m_2 = 5k + 3k = 8k$$
Substitute this into the sum formula:
$$8k = -\frac{2h}{b} \implies k = -\frac{2h}{8b} = -\frac{h}{4b} \quad \dots \text{(Equation 1)}$$
2. Find the product of the slopes in terms of $k$:
$$m_1m_2 = (5k)(3k) = 15k^2$$
Substitute this into the product formula:
$$15k^2 = \frac{a}{b} \implies k^2 = \frac{a}{15b} \quad \dots \text{(Equation 2)}$$
3. Substitute the value of $k$ from Equation 1 into Equation 2:
$$\left(-\frac{h}{4b}\right)^2 = \frac{a}{15b}$$
$$\frac{h^2}{16b^2} = \frac{a}{15b}$$
Cancel out one factor of $b$ from both denominators:
$$\frac{h^2}{16b} = \frac{a}{15}$$
Cross-multiply to isolate the ratio $\frac{h^2}{ab}$:
$$15h^2 = 16ab \implies \frac{h^2}{ab} = \frac{16}{15}$$
Therefore, the ratio $h^2 : ab$ is equal to $16 : 15$, which corresponds perfectly to option (D).
Step 4: Final Answer:
The ratio $h^2 : ab$ is $16 : 15$, which corresponds to option (D).