question_answer

If the straight line \[x\text{ }=\text{ }y\sqrt{3,}\] cuts the ellipse \[{{x}^{2}}+{{y}^{2}}+xy=3\] at points P and Q, then \[\left| \,OP\, \right|\,\,\,\left| \,OQ\, \right|\] 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).

\[\because \,\,OP=OQ=r\]

\[\because \] 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)

\[\Rightarrow \frac{3{{r}^{2}}}{4}+\frac{{{r}^{2}}}{4}+\frac{{{r}^{2}}\sqrt{3}}{4}-3=0\]\[\Rightarrow {{r}^{2}}=\frac{3\times 4}{(4+\sqrt{3})}\]\[\Rightarrow {{r}^{2}}=\frac{12}{4+\sqrt{3}}\]

\[\therefore \left| \,OP\, \right|\,\,\,\left| \,OQ\, \right|=\frac{12}{4+\sqrt{3}}\]