Question:

The equation of bisectors of the angle between the lines given by \( 3x^2 + 5xy + 4y^2 = 0 \) is:

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Angle Bisectors of Homogeneous Second-Degree Equations}
Use standard formula: \( x^2 - y^2 = \frac{a - b}{h}xy \)
Ensure signs are handled correctly during simplification
These bisectors pass through the origin and divide the angle between the lines
Updated On: May 19, 2025
  • \( x^2 - y^2 - \frac{2}{5}xy = 0 \)
  • \( x^2 + y^2 + \frac{2}{5}xy = 0 \)
  • \( x^2 - y^2 - \frac{1}{5}xy = 0 \)
  • \( x^2 - y^2 + \frac{1}{5}xy = 0 \)
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The Correct Option is A

Solution and Explanation

The equation \( 3x^2 + 5xy + 4y^2 = 0 \) represents a pair of straight lines passing through the origin. To find the angle bisectors of such lines: - For general form \( ax^2 + 2hxy + by^2 = 0 \), the angle bisectors are given by: \[ \frac{x^2 - y^2}{a - b} = \frac{xy}{h} \Rightarrow x^2 - y^2 = \frac{a - b}{h}xy \] Here, \( a = 3, h = \frac{5}{2}, b = 4 \) \[ x^2 - y^2 = \frac{3 - 4}{5/2}xy = \left( -\frac{1}{5/2} \right) xy = -\frac{2}{5}xy \Rightarrow x^2 - y^2 + \frac{2}{5}xy = 0 \] Move term to RHS: \[ x^2 - y^2 - \frac{2}{5}xy = 0 \]
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