Question 45

Mathematics Line Easy

If one of the lines of \( a x^2+2 h x y+b y^2=0 \) bisects the angle between the axes in the first quadrant, then

(A) \( h^2-a b=0 \)
(B) \( h^2+a b=0 \)
(C) \( (a+b)^2=h^2 \)
(D) \( (a+b)^2=4 h^2 \)
View Dynamic Solution & Explanation
Correct Solution: Option D

Step-by-step Solution:

Step 1: Substitute \( y = x \) into the Equation Substitute \( y = x \) into \( ax^2 + 2hxy + by^2 = 0 \): \[ ax^2 + 2hx \cdot x + bx^2 = 0 \] \[ ax^2 + 2hx^2 + bx^2 = 0 \] \[ (a + 2h + b)x^2 = 0 \] Since this must hold for all \( x \), the coefficient of \( x^2 \) must be zero: \[ a + 2h + b = 0 \] Step 2: Solve for \( a + b \) Rearrange the equation: \[ a + b = -2h \] Step 3: Square Both Sides Square both sides to find a relationship between \( a + b \) and \( h \): \[ (a + b)^2 = (-2h)^2 \] \[ (a + b)^2 = 4h^2 \] Step 4: Conclusion The relationship between \( a + b \) and \( h \) is: \[ \boxed{(a + b)^2 = 4h^2} \]