A) \[a{{x}^{2}}-2hxy+b{{y}^{2}}=0\]
B) \[b{{x}^{2}}+2hxy+a{{y}^{2}}=0\]
C) \[a{{y}^{2}}-2hxy+b{{x}^{2}}=0\]
D) \[a{{y}^{2}}-b{{x}^{2}}=0\]
Correct Answer: C
Solution :
Perpendicular lines are given by \[b{{x}^{2}}-2hxy+a{{y}^{2}}=0\].
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