Question:
If the angle made by the lines represented by the equation \(ax^2+2hxy+by^2 = 0\) with X-axis are \(α\) and \(β\), then \(tan(α+β)\) is
If the angle made by the lines represented by the equation \(ax^2+2hxy+by^2 = 0\) with X-axis are \(α\) and \(β\), then \(tan(α+β)\) is
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The slopes tan(alpha) and tan(beta) are roots of b m^2 + 2h m + a = 0.
Updated On: Oct 1, 2026
- \(\frac{h}{a+b}\)
- \(\frac{2h}{a-b}\)
- \(\frac{2h}{a+b}\)
- \(\frac{h}{a-b}\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
The lines given by \(ax^2+2hxy+by^2=0\) pass through the origin. If they make angles \(\alpha\) and \(\beta\) with the X-axis, their slopes are \(m_1=\tan\alpha\) and \(m_2=\tan\beta\).
Step 2: Slope equation
Put \(y=mx\) in the equation:
\[ a+2hm+bm^2=0 \]
\[ m_1+m_2=-\frac{2h}{b},\qquad m_1m_2=\frac{a}{b} \]
Step 3: Tangent of the sum
\[ \tan(\alpha+\beta)=\frac{m_1+m_2}{1-m_1m_2}=\frac{-2h/b}{1-a/b} \]
\[ =\frac{-2h}{b-a}=\frac{2h}{a-b} \]
Step 4: Check
Option (B) is exactly this. Options (A) and (D) miss the factor 2 and option (C) has a plus sign in the denominator.
Final Answer:
\(\tan(\alpha+\beta)=\dfrac{2h}{a-b}\), option (B). \[ \boxed{\frac{2h}{a-b}} \]
The lines given by \(ax^2+2hxy+by^2=0\) pass through the origin. If they make angles \(\alpha\) and \(\beta\) with the X-axis, their slopes are \(m_1=\tan\alpha\) and \(m_2=\tan\beta\).
Step 2: Slope equation
Put \(y=mx\) in the equation:
\[ a+2hm+bm^2=0 \]
\[ m_1+m_2=-\frac{2h}{b},\qquad m_1m_2=\frac{a}{b} \]
Step 3: Tangent of the sum
\[ \tan(\alpha+\beta)=\frac{m_1+m_2}{1-m_1m_2}=\frac{-2h/b}{1-a/b} \]
\[ =\frac{-2h}{b-a}=\frac{2h}{a-b} \]
Step 4: Check
Option (B) is exactly this. Options (A) and (D) miss the factor 2 and option (C) has a plus sign in the denominator.
Final Answer:
\(\tan(\alpha+\beta)=\dfrac{2h}{a-b}\), option (B). \[ \boxed{\frac{2h}{a-b}} \]
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