Question

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 OA,OB,OC,OD (where O is the origin) is

• A. $a-2b+c$
• B. $2c^{2}d$
• C. $96$
• D. $6$

Answer 1

The correct option is C $96$

$y=\sqrt{3}x$

$\frac{y}{\sin 60}=\frac{x}{\cos 60}=r$

$(x,y)=(\frac{r}{2},\frac{\sqrt{3}r}{2})...\left(i\right)$

$x^4+a{x}^{2}y+6xy+cx+dy+6=\mathrm{0...}\left(ii\right)$

Put (i) in (ii)

${\left(\frac{r}{2}\right)}^4+a{\left(\frac{r}{2}\right)}^2\left(\frac{\sqrt{3}r}{2}\right)+6\left(\frac{r}{2}\right)\left(\frac{\sqrt{3}r}{2}\right)+c\left(\frac{r}{2}\right)+d\left(\frac{\sqrt{3}r}{2}\right)+6=0$

$(\frac{r^4}{16}+\frac{\sqrt{3}{r}^3}{2}+\frac{\sqrt{3}{r}^2}{2}+\frac{(c+d\sqrt{3})r}{2}+6=0)\times 16$

$r^4+2a\sqrt{3}{r}^3+4\sqrt{3}6r^2+(c+d\sqrt{3})8r+96=\mathrm{0.....}\left(iii\right)$

Now as $y=\sqrt{3}x$ cut (ii) at $A,B,C,D$

Let $OA=r_1,OB=r_2,OC=r_3,OD=r_4$

$OA.OB.OC.OD=r_1.r_2.r_3.r_4$ (From (iii))

hence $OA.OB.OC.OD=96$