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JEE Main 2019Medium

Q.The pair of lines ax2+2hxy+by2=0ax^2 + 2hxy + by^2 = 0 bisects the angle between the pair a′x2+2h′xy+b′y2=0a'x^2 + 2h'xy + b'y^2 = 0 if

a
a+bh=a′+b′h′\dfrac{a + b}{h} = \dfrac{a' + b'}{h'}
b
a−bh=a′−b′h′\dfrac{a - b}{h} = \dfrac{a' - b'}{h'}
c
ah+bh=a′h′+b′h′\dfrac{a}{h} + \dfrac{b}{h} = \dfrac{a'}{h'} + \dfrac{b'}{h'}
d
ah−bh=a′h′−b′h′\dfrac{a}{h} - \dfrac{b}{h} = \dfrac{a'}{h'} - \dfrac{b'}{h'}

Correct Answer: Option A

The correct solution involves applying the fundamental concept to derive the final value step by step...

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