A) 6
B) 8
C) 12
D) 16
Correct Answer: B
Solution :
We have, \[y=\sqrt{3}x\] |
Let \[x=\frac{r}{2},y=\frac{\sqrt{3}r}{2}\] |
Putting the value of x and y in \[{{x}^{4}}+a{{x}^{2}}y+bxy+cx+dy+6=0,\] we get |
\[\frac{{{r}^{4}}}{16}+\frac{\sqrt{3}a{{r}^{3}}}{8}+\frac{\sqrt{3}b{{r}^{2}}}{4}+\frac{cr}{2}+\frac{\sqrt{3}dr}{2}+6=0\] |
\[\Rightarrow \]\[\frac{1}{12}(OA\cdot OB\cdot OC\cdot OD)=\frac{1}{12}\left| {{r}_{1}}{{r}_{2}}{{r}_{3}}{{r}_{4}} \right|\]\[=\frac{1}{12}(16\times 6)=\frac{96}{12}=8\] |
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