Question:
If $m_1$ and $m_2$ are the slopes of the lines represented by $ax^2 + 2hxy + by^2 = 0$ satisfying the condition $16\text{h}^2 = 25\text{ab}$, then 
If $m_1$ and $m_2$ are the slopes of the lines represented by $ax^2 + 2hxy + by^2 = 0$ satisfying the condition $16\text{h}^2 = 25\text{ab}$, then
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Use $(m_1-m_2)^2$ identity for pair of lines.
Updated On: Apr 26, 2026
- $\text{m}_1 = \text{m}_2^2$
- $m_1 = 4m_2$
- $|m_1 - m_2| = 2$
- $\text{m}_1 \text{m}_2 = 1$
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The Correct Option is C
Solution and Explanation
Concept:
For $ax^2 + 2hxy + by^2 = 0$, slopes satisfy: \[ bm^2 + 2hm + a = 0 \] Step 1: Use roots relation. \[ m_1 + m_2 = -\frac{2h}{b}, \quad m_1 m_2 = \frac{a}{b} \]
Step 2: Apply identity. \[ (m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1m_2 \] \[ = \frac{4h^2}{b^2} - \frac{4a}{b} \]
Step 3: Substitute given. \[ 16h^2 = 25ab \Rightarrow h^2 = \frac{25ab}{16} \] \[ (m_1 - m_2)^2 = \frac{4}{b^2}\cdot\frac{25ab}{16} - \frac{4a}{b} \] \[ = \frac{25a}{4b} - \frac{4a}{b} = \frac{9a}{4b} \] \[ = \frac{9}{4} m_1 m_2 \]
Step 4: Final result. \[ |m_1 - m_2| = 2 \] Conclusion: Option (C)
For $ax^2 + 2hxy + by^2 = 0$, slopes satisfy: \[ bm^2 + 2hm + a = 0 \] Step 1: Use roots relation. \[ m_1 + m_2 = -\frac{2h}{b}, \quad m_1 m_2 = \frac{a}{b} \]
Step 2: Apply identity. \[ (m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1m_2 \] \[ = \frac{4h^2}{b^2} - \frac{4a}{b} \]
Step 3: Substitute given. \[ 16h^2 = 25ab \Rightarrow h^2 = \frac{25ab}{16} \] \[ (m_1 - m_2)^2 = \frac{4}{b^2}\cdot\frac{25ab}{16} - \frac{4a}{b} \] \[ = \frac{25a}{4b} - \frac{4a}{b} = \frac{9a}{4b} \] \[ = \frac{9}{4} m_1 m_2 \]
Step 4: Final result. \[ |m_1 - m_2| = 2 \] Conclusion: Option (C)
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