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  • If the line cuts the curve <img alt="x^3+a x^2+b x-72=0" src="https://entrancecorner.oncodeco

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

 

Option: 1

576


Option: 2

-576


Option: 3

a+b-c-576


Option: 4

a+b+c-576


Answers (1)

best_answer

The line y=\sqrt{3} x is passing through the origin and slope is \sqrt{3}, hence in parametric form the equation of given line can be written as
                        \frac{x}{1 / 2}=\frac{y}{\sqrt{3} / 2}=r
Any point on the line (i) is \left(\frac{r}{2}, \frac{\sqrt{3} r}{2}\right). If the line cuts the given curve, then
                    \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 r_1, I_2, r_3  will represent O A, O B, O C.
Therefore,  O A \cdot O B \cdot OC=\left|r_1\right|\left|I_2\right|\left|r_3\right|=576 

 

Posted by

Ritika Jonwal

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