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KVPY Sample Paper KVPY Stream-SX Model Paper-13

  • question_answer
    If the line \[y=\sqrt{3}x\] cuts the curve \[{{x}^{4}}+a{{x}^{2}}y+bxy+cx+dy+6=0\] at A, B, C and D, then the value of \[\frac{1}{12}OA\cdot OB\cdot OC\cdot OD\](where O is origin) is equal to

    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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