A) \[\frac{1}{5}\]
B) 1
C) \[\frac{7}{5}\]
D) 7
Correct Answer: A
Solution :
[a] We have, \[2{{x}^{2}}+5xy+3{{y}^{2}}+6x+7y+4=0\] comparing this eq. with \[a{{x}^{2}}+b{{y}^{2}}+2hxy+2gx+2fy+c=0.\] we get \[a=2,b=3,h=\frac{5}{2}\] \[\therefore \tan \theta =\frac{2\sqrt{{{h}^{2}}-ab}}{a+b}=\frac{2\sqrt{\frac{25}{4}-2\times 3}}{2+3}\] \[=\frac{2\sqrt{\frac{1}{4}}}{5}=\frac{2\times \frac{1}{2}}{5}=\frac{1}{5}\tan \theta =\frac{1}{5}\Rightarrow m=\frac{1}{5}\]
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