If the line y = sqrt{3}x cuts the curve x^4 + | vedclass

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(C) Any point on the line $y = \sqrt{3}x$ can be represented in polar form as $(r \cos 	heta, r \sin 	heta)$,where $	heta = 60^\circ = \frac{\pi}{3

Vedclass · Accepted Answer

$96$

Vedclass · Answer

(C) Any point on the line $y = \sqrt{3}x$ can be represented in polar form as $(r \cos 	heta, r \sin 	heta)$,where $	heta = 60^\circ = \frac{\pi}{3}$.Thus,the coordinates are $\left(r \cos 60^\circ, r \sin 60^\circight) = \left(\frac{r}{2}, \frac{r\sqrt{3}}{2}ight)$.Substituting these into the curve equation $x^4 + ax^2y + bxy + cx + dy + 6 = 0$:$\left(\frac{r}{2}ight)^4 + a\left(\frac{r}{2}ight)^2\left(\frac{r\sqrt{3}}{2}ight) + b\left(\frac{r}{2}ight)\left(\frac{r\sqrt{3}}{2}ight) + c\left(\frac{r}{2}ight) + d\left(\frac{r\sqrt{3}}{2}ight) + 6 = 0$$\frac{r^4}{16} + \frac{a\sqrt{3}r^3}{8} + \frac{b\sqrt{3}r^2}{4} + \frac{r(c + d\sqrt{3})}{2} + 6 = 0$Multiplying by $16$:$r^4 + (2a\sqrt{3})r^3 + (4b\sqrt{3})r^2 + 8(c + d\sqrt{3})r + 96 = 0$This is a quartic equation in $r$ where the roots $r_1, r_2, r_3, r_4$ represent the distances $OA, OB, OC, OD$ from the origin.By Vieta's formulas,the product of the roots is the constant term divided by the leading coefficient:$OA \cdot OB \cdot OC \cdot OD = r_1 r_2 r_3 r_4 = \frac{96}{1} = 96$.

If the line $y = \sqrt{3}x$ cuts the curve $x^4 + ax^2y + bxy + cx + dy + 6 = 0$ at $A$,$B$,$C$,and $D$,then the value of $OA \cdot OB \cdot OC \cdot OD$ is,(where $O$ is the origin).

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