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  • If the line cuts the curve <img alt="\mathrm{x^{3}+a x^{2}+b x-72=0}"

If the line \mathrm{y=\sqrt{3} x} cuts the curve \mathrm{x^{3}+a x^{2}+b x-72=0}  at A, B and C, then OA. OB.OC (Where ' O ' is origin) is

Option: 1

576


Option: 2

-576


Option: 3

\mathrm{a+b-c-576}


Option: 4

\mathrm{a+b+c-576}


Answers (1)

best_answer

The line \mathrm{y=\sqrt{3} x} is passing through the origin and slope is \mathrm{\sqrt{3}}, hence in parametric form the equation of given line can be written as

\mathrm{\frac{x}{1 / 2}=\frac{y}{\sqrt{3} / 2}=r \quad \ldots (1) }

Any point on the line (1) is \mathrm{\left(\frac{r}{2}, \frac{\sqrt{3} r}{2}\right) }.If the line cuts the given curve, then \mathrm{\frac{r^{3}}{8}+\frac{a r^{2}}{4}+\frac{b r}{2}-72=0}.This is a cubic equation in r. Roots of this equation \mathrm{r_{1}, r_{2}, r_{3}} will
represent OA, OB and OC.
Therefore OA.OB.OC \mathrm{=\left|r_{1}\right|\left|r_{2}\right|\left|r_{3}\right|=576}

Hence (A) is the correct answer.

Posted by

Devendra Khairwa

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