Question

If $4ab = 3h^2$, then the ratio of slopes of the lines represented by $ax^2 + 2hxy + by^2 = 0$ is

• A. $2 : 1$
• B. $\sqrt{2} : 1$
• C. $3 : 1$
• D. $1 : 3$

Hint:

Here is a very helpful general formula shortcut for calculating slope ratios: if the slopes of the lines $ax^2+2hxy+by^2=0$ are in the ratio $1 : n$, then the relationship between the coefficients is always given by $\frac{(n+1)^2}{n} = \frac{4h^2}{ab}$. Substituting $n=3$ gives $\frac{(3+1)^2}{3} = \frac{16}{3} = \frac{4h^2}{ab} \implies 4ab = 3h^2$, proving it works perfectly!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given a condition connecting the coefficients of a pair of straight lines, $4ab = 3h^2$. We need to determine the numerical ratio of their individual slopes, $m_1 : m_2$.

Step 2: Key Formula or Approach:
For a standard homogenous second-degree pair of lines $ax^2 + 2hxy + by^2 = 0$, the sum and product of the slopes are:
$$m_1 + m_2 = -\frac{2h}{b} \quad \text{and} \quad m_1m_2 = \frac{a}{b}$$ We can determine the difference of the slopes using the classic algebraic identity:
$$(m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1m_2$$

Step 3: Detailed Explanation:
Let's substitute the sum and product formulas into the identity:
$$(m_1 - m_2)^2 = \left(-\frac{2h}{b}\right)^2 - 4\left(\frac{a}{b}\right) = \frac{4h^2}{b^2} - \frac{4a}{b} = \frac{4h^2 - 4ab}{b^2}$$ We are given the condition $4ab = 3h^2 \implies -4ab = -3h^2$. Let's substitute this value into the numerator:
$$(m_1 - m_2)^2 = \frac{4h^2 - 3h^2}{b^2} = \frac{h^2}{b^2}$$ Taking the square root of both sides gives:
$$m_1 - m_2 = \frac{h}{b}$$ Now we can solve for $m_1$ and $m_2$ by setting up a system of two linear equations using our sum and difference expressions:
1. $m_1 + m_2 = -\frac{2h}{b}$
2. $m_1 - m_2 = \frac{h}{b}$
Adding the two equations together:
$$2m_1 = -\frac{2h}{b} + \frac{h}{b} = -\frac{h}{b} \implies m_1 = -\frac{h}{2b}$$ Subtracting the second equation from the first equation:
$$2m_2 = -\frac{2h}{b} - \frac{h}{b} = -\frac{3h}{b} \implies m_2 = -\frac{3h}{2b}$$ Now compute the final ratio of the two slopes:
$$\frac{m_1}{m_2} = \frac{-\frac{h}{2b}}{-\frac{3h}{2b}} = \frac{1}{3} \implies m_1 : m_2 = 1 : 3$$ This matches option (D).

Step 4: Final Answer:
The ratio of the slopes of the lines is $1 : 3$, which corresponds to option (D).