question_answer

If the straight line \[x=y\sqrt{3}\] cuts the ellipse \[{{x}^{2}}+{{y}^{2}}+xy=3\] at points P and Q, then |OP| |OQ| is (where 'O' is the origin)

A) \[4+\sqrt{3}\]               

B) \[\frac{12}{4-\sqrt{3}}\]

C) \[\frac{12}{4+\sqrt{3}}\]           

D) \[\frac{-12}{4+\sqrt{3}}\]

Correct Answer: C

Solution :

\[{{x}^{2}}+{{y}^{2}}+xy-3=0\]		? (1)
\[y=\frac{x}{\sqrt{3}}\]		? (2)
\[\because \]       centre of ellipse is (0, 0) and (2) also passes through (0, 0)
\[\therefore \]      chord will be bisect at (0, 0).
\[\because \]       Equation PQ is\[\frac{x-0}{\cos \frac{\pi }{6}}=\frac{y-0}{\sin \frac{\pi }{6}}=\pm \,r.\]
\[\therefore \]      \[\left( \frac{r\sqrt{3}}{2},\,\,\frac{r}{2} \right)\] lies on	? (1)
\[\Rightarrow \]   \[\frac{3{{r}^{2}}}{4}+\frac{{{r}^{2}}}{4}+\frac{{{r}^{2}}\sqrt{3}}{4}-3=0\]
\[\Rightarrow \]   \[{{r}^{2}}=\frac{3\times 4}{(4+\sqrt{3})}\]     \[\Rightarrow \]            \[{{r}^{2}}=\frac{12}{4+\sqrt{3}}\]
\[\therefore \]      \[|OP||OQ|\,\,=\frac{12}{4+\sqrt{3}}\]