Question

If the line $y=\sqrt{3}x$ cuts the curve ${{x}^{3}}+a{{x}^{2}}+bx-72=0$ at $A,B$ and $C$, the $OA\times OB\times OC$ (where $O$ is origin) is:
A. $576$
B. $-576$
C. $a+b-c-576$
D. $a+b+c-576$

Answer 1

Hint: To solve this question, we will find the value of $x$ from the equation of line $y=\sqrt{3}x$ that will be $x=\dfrac{y}{\sqrt{3}}$. Then, we will substitute the value of $x$ in the equation of curve ${{x}^{3}}+a{{x}^{2}}+bx-72=0$ and will find the relation between its roots that will be also the relation between coordinates of $A,B$ and $C$. Then, we will use $OA\times OB\times OC$ 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:
$\Rightarrow y=\sqrt{3}x$
After simplifying it, we will get the value of $x$ as:
$x=\dfrac{y}{\sqrt{3}}$
Now, we will use $x=\dfrac{y}{\sqrt{3}}$ in the equation of curve as:
$\Rightarrow {{\left( \dfrac{y}{\sqrt{3}} \right)}^{3}}+a{{\left( \dfrac{y}{\sqrt{3}} \right)}^{2}}+b\left( \dfrac{y}{\sqrt{3}} \right)-72=0$
Here, we will simplify the above expression as:
\[\Rightarrow \dfrac{{{y}^{3}}}{3\sqrt{3}}+a\dfrac{{{y}^{2}}}{3}+b\dfrac{y}{\sqrt{3}}-72=0\]
Now, we will multiply with $3\sqrt{3}$ in the above equation and will simplify it as:
\[\begin{align}
  & \Rightarrow 3\sqrt{3}\times \dfrac{{{y}^{3}}}{3\sqrt{3}}+3\sqrt{3}\times \dfrac{a{{y}^{2}}}{3}+3\sqrt{3}\times \dfrac{by}{\sqrt{3}}-3\sqrt{3}\times 72=0 \\
 & \Rightarrow {{y}^{3}}+\sqrt{3}a{{y}^{2}}+3by-216\sqrt{3}=0 \\
\end{align}\]
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:
$\begin{align}
  & \Rightarrow \sqrt{\dfrac{{{y}_{1}}^{2}}{3}+{{y}_{1}}^{2}} \\
 & \Rightarrow \sqrt{\dfrac{{{y}_{1}}^{2}+3{{y}_{1}}^{2}}{3}} \\
 & \Rightarrow \sqrt{\dfrac{4{{y}_{1}}^{2}}{3}} \\
 & \Rightarrow \dfrac{2{{y}_{1}}}{\sqrt{3}} \\
\end{align}$
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.
\[\begin{align}
  & \Rightarrow OA\times OB\times OC=\dfrac{8}{3}\times 216 \\
 & \Rightarrow OA\times OB\times OC=8\times 72 \\
 & \Rightarrow OA\times OB\times OC=576 \\
\end{align}\]
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:
$\begin{align}
  & OB= \sqrt{\dfrac{{{y}_{2}}^{2}}{3}+{{y}_{2}}^{2}} \\
 & OB= \sqrt{\dfrac{{{y}_{2}}^{2}+3{{y}_{2}}^{2}}{3}} \\
 & OB= \sqrt{\dfrac{4{{y}_{2}}^{2}}{3}} \\
 & OB= \dfrac{2{{y}_{2}}}{\sqrt{3}} \\
\end{align}$
$OC= \sqrt{{{\left( {{x}_{3}}-0 \right)}^{2}}+{{\left( {{y}_{3}}-0 \right)}^{2}}}$
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:
$\begin{align}
  & OC= \sqrt{\dfrac{{{y}_{3}}^{2}}{3}+{{y}_{3}}^{2}} \\
 & OC= \sqrt{\dfrac{{{y}_{3}}^{2}+3{{y}_{3}}^{2}}{3}} \\
 & OC= \sqrt{\dfrac{4{{y}_{3}}^{2}}{3}} \\
 & OC= \dfrac{2{{y}_{3}}}{\sqrt{3}} \\
\end{align}$