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Question: If the line \(y=\sqrt{3}x\) cuts the curve \({{x}^{3}}+a{{x}^{2}}+bx-72=0\) at \(A,B\) and \(C\), th...

If the line y=3xy=\sqrt{3}x cuts the curve x3+ax2+bx72=0{{x}^{3}}+a{{x}^{2}}+bx-72=0 at A,BA,B and CC, the OA×OB×OCOA\times OB\times OC (where OO is origin) is:
A. 576576
B. 576-576
C. a+bc576a+b-c-576
D. a+b+c576a+b+c-576

Explanation

Solution

To solve this question, we will find the value of xx from the equation of line y=3xy=\sqrt{3}x that will be x=y3x=\dfrac{y}{\sqrt{3}}. Then, we will substitute the value of xx in the equation of curve x3+ax2+bx72=0{{x}^{3}}+a{{x}^{2}}+bx-72=0 and will find the relation between its roots that will be also the relation between coordinates of A,BA,B and CC. Then, we will use OA×OB×OCOA\times OB\times OC and substitute the corresponding values to simplify it and will find the required answer.

Complete step-by-step solution:
Since, given that the equation of line:
y=3x\Rightarrow y=\sqrt{3}x
After simplifying it, we will get the value of xx as:
x=y3x=\dfrac{y}{\sqrt{3}}
Now, we will use x=y3x=\dfrac{y}{\sqrt{3}} in the equation of curve as:
(y3)3+a(y3)2+b(y3)72=0\Rightarrow {{\left( \dfrac{y}{\sqrt{3}} \right)}^{3}}+a{{\left( \dfrac{y}{\sqrt{3}} \right)}^{2}}+b\left( \dfrac{y}{\sqrt{3}} \right)-72=0
Here, we will simplify the above expression as:
y333+ay23+by372=0\Rightarrow \dfrac{{{y}^{3}}}{3\sqrt{3}}+a\dfrac{{{y}^{2}}}{3}+b\dfrac{y}{\sqrt{3}}-72=0
Now, we will multiply with 333\sqrt{3} in the above equation and will simplify it as:

& \Rightarrow 3\sqrt{3}\times \dfrac{{{y}^{3}}}{3\sqrt{3}}+3\sqrt{3}\times \dfrac{a{{y}^{2}}}{3}+3\sqrt{3}\times \dfrac{by}{\sqrt{3}}-3\sqrt{3}\times 72=0 \\\ & \Rightarrow {{y}^{3}}+\sqrt{3}a{{y}^{2}}+3by-216\sqrt{3}=0 \\\ \end{aligned}$$ Since, the degree of the above polynomial is $3$. Then, there must be three roots of the above equation. Let’s consider ${{y}_{1}},{{y}_{2}}$ and ${{y}_{3}}$. So, we will get some relation as: $\Rightarrow {{y}_{1}}+{{y}_{2}}+{{y}_{3}}=-\sqrt{3}a$ $\Rightarrow {{y}_{1}}{{y}_{2}}+{{y}_{2}}{{y}_{3}}+{{y}_{3}}{{y}_{1}}=3b$ $\Rightarrow {{y}_{1}}\cdot {{y}_{2}}\cdot {{y}_{3}}=-\left( -216\sqrt{3} \right)$ It can be written as: $\Rightarrow {{y}_{1}}\cdot {{y}_{2}}\cdot {{y}_{3}}=216\sqrt{3}$ Since, point $O$ is the origin. Then, the point of $O$ is $$\left( 0,0 \right)$$. Now, let’s suppose that the points of $A,B$ and $C$ are $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)$ and $\left( {{x}_{3}},{{y}_{3}} \right)$ So, the value of line $OA$is: $\Rightarrow \sqrt{{{\left( {{x}_{1}}-0 \right)}^{2}}+{{\left( {{y}_{1}}-0 \right)}^{2}}}$ Now, we will simplify it as: $\Rightarrow \sqrt{{{x}_{1}}^{2}+{{y}_{1}}^{2}}$ From the equation of line, we will have ${{x}_{1}}=\dfrac{{{y}_{1}}}{\sqrt{3}}$. We will use it in the above step as: $\Rightarrow \sqrt{{{\left( \dfrac{{{y}_{1}}}{\sqrt{3}} \right)}^{2}}+{{y}_{1}}^{2}}$ Now, we will simplify it as: $\begin{aligned} & \Rightarrow \sqrt{\dfrac{{{y}_{1}}^{2}}{3}+{{y}_{1}}^{2}} \\\ & \Rightarrow \sqrt{\dfrac{{{y}_{1}}^{2}+3{{y}_{1}}^{2}}{3}} \\\ & \Rightarrow \sqrt{\dfrac{4{{y}_{1}}^{2}}{3}} \\\ & \Rightarrow \dfrac{2{{y}_{1}}}{\sqrt{3}} \\\ \end{aligned}$ Here, we find the value of $OA$ as $\dfrac{2{{y}_{1}}}{\sqrt{3}}$. Similarly, we will get the value of $OB$ and $OC$ as $\dfrac{2{{y}_{2}}}{\sqrt{3}}$ and $\dfrac{2{{y}_{3}}}{\sqrt{3}}$. Now, we will need to find the value of $OA\times OB\times OC$. So, we will substitute the respective values as: $$\Rightarrow OA\times OB\times OC=\dfrac{2{{y}_{1}}}{\sqrt{3}}\times \dfrac{2{{y}_{2}}}{\sqrt{3}}\times \dfrac{2{{y}_{2}}}{\sqrt{3}}$$ Here, we will simplify it as: $$\Rightarrow OA\times OB\times OC=\dfrac{8}{3\sqrt{3}}\times {{y}_{1}}{{y}_{2}}{{y}_{3}}$$ Now, we will substitute $$216\sqrt{3}$$ for ${{y}_{1}}{{y}_{2}}{{y}_{3}}$ as we got from the multiple of roots of equation $${{y}^{3}}+\sqrt{3}a{{y}^{2}}+3by-216\sqrt{3}=0$$. So, we will have the above equation as: $$\Rightarrow OA\times OB\times OC=\dfrac{8}{3\sqrt{3}}\times 216\sqrt{3}$$ After simplification, we will get the required answer. $$\begin{aligned} & \Rightarrow OA\times OB\times OC=\dfrac{8}{3}\times 216 \\\ & \Rightarrow OA\times OB\times OC=8\times 72 \\\ & \Rightarrow OA\times OB\times OC=576 \\\ \end{aligned}$$ **Hence, option $A$ is correct.** **Note:** Here, we will do the calculation for $OB$ and $OC$ as: $OB= \sqrt{{{\left( {{x}_{2}}-0 \right)}^{2}}+{{\left( {{y}_{2}}-0 \right)}^{2}}}$ Now, we will simplify it as: $OB= \sqrt{{{x}_{2}}^{2}+{{y}_{2}}^{2}}$ From the equation of the line, we will have ${{x}_{2}}=\dfrac{{{y}_{2}}}{\sqrt{3}}$. We will use it in the above step as: $OB= \sqrt{{{\left( \dfrac{{{y}_{2}}}{\sqrt{3}} \right)}^{2}}+{{y}_{2}}^{2}}$ Now, we will simplify it as: $\begin{aligned} & OB= \sqrt{\dfrac{{{y}_{2}}^{2}}{3}+{{y}_{2}}^{2}} \\\ & OB= \sqrt{\dfrac{{{y}_{2}}^{2}+3{{y}_{2}}^{2}}{3}} \\\ & OB= \sqrt{\dfrac{4{{y}_{2}}^{2}}{3}} \\\ & OB= \dfrac{2{{y}_{2}}}{\sqrt{3}} \\\ \end{aligned}$ $OC= \sqrt{{{\left( {{x}_{3}}-0 \right)}^{2}}+{{\left( {{y}_{3}}-0 \right)}^{2}}}$ Now, we will simplify it as: $OC= \sqrt{{{x}_{3}}^{2}+{{y}_{3}}^{2}}$ From the equation of line, we will have ${{x}_{3}}=\dfrac{{{y}_{3}}}{\sqrt{3}}$. We will use it in the above step as: $OC= \sqrt{{{\left( \dfrac{{{y}_{3}}}{\sqrt{3}} \right)}^{2}}+{{y}_{3}}^{2}}$ Now, we will simplify it as: $\begin{aligned} & OC= \sqrt{\dfrac{{{y}_{3}}^{2}}{3}+{{y}_{3}}^{2}} \\\ & OC= \sqrt{\dfrac{{{y}_{3}}^{2}+3{{y}_{3}}^{2}}{3}} \\\ & OC= \sqrt{\dfrac{4{{y}_{3}}^{2}}{3}} \\\ & OC= \dfrac{2{{y}_{3}}}{\sqrt{3}} \\\ \end{aligned}$