A) \[3x+3y=40\]
B) \[3x+3y+40=0\]
C) \[3x-3y=40\]
D) \[3y-3x=40\]
Correct Answer: D
Solution :
\[x=r\,\cos \,\theta ,\,\,y=r\sin \theta \] \[\therefore \]putting in \[{{L}_{1}}\], we get \[\frac{1}{OA}=\frac{\sin \theta -\cos \theta }{10}\] Similarly, putting in \[{{L}_{2}}\], we get \[\frac{1}{OB}=\frac{\sin \,\theta -\cos \theta }{20}\] \[\Rightarrow \,h=r\cos \theta ,k=r\sin \theta \] \[\therefore \,\frac{2}{r}=\left( \frac{\sin \theta -\cos \theta }{10} \right)+\left( \frac{\sin \theta -\cos \theta }{20} \right)\] \[\Rightarrow 40=3(r\,\sin \,\theta )-3(r\,\cos \,\theta )\,\Rightarrow \,3y-3x=40\]You need to login to perform this action.
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