A) a
B) b
C) \[h\]
D) Any real number
Correct Answer: D
Solution :
Bisectors of \[a{{x}^{2}}+2hxy+b{{y}^{2}}=0\] are \[\frac{{{x}^{2}}-{{y}^{2}}}{a-b}=\frac{xy}{h}\] .....(i) and of \[a{{x}^{2}}+2hxy+b{{y}^{2}}+\lambda ({{x}^{2}}+{{y}^{2}})=0\] i.e., \[(a+\lambda ){{x}^{2}}+2hxy+(b+\lambda ){{y}^{2}}=0\]are \[\frac{{{x}^{2}}-{{y}^{2}}}{(a+\lambda )-(b+\lambda )}=\frac{xy}{h}\] .....(ii) Which is the same equation as equation (i). Hence for any \[\lambda \]belonging to real numbers, the lines will have same bisectors.
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