Question

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. $\frac{-12}{4+\sqrt{3}}$

Answer 1

The correct option is C $\frac{12}{4+\sqrt{3}}$

$x=y\sqrt{3}$ cuts the ellipse $x^2+y^2+xy=3$

${3y}^2+y^2+\sqrt{3}{y}^2=3$

$y=\pm \sqrt{\frac{3}{\sqrt{3}+4}}andx=\pm \sqrt{3}\sqrt{\frac{3}{\sqrt{3}+4}}$

Now,

$\left|OP\right|\times \left|OQ\right|={\left[\sqrt{{\left(\sqrt{3}\sqrt{\frac{3}{\sqrt{3}+4}}\right)}^2+{\left(\sqrt{\frac{3}{\sqrt{3}+4}}\right)}^2}\right]}^2$

$\left|OP\right|\times \left|OQ\right|=3\times \frac{3}{\sqrt{3}+4}+\frac{3}{\sqrt{3}+4}$

$\left|OP\right|\times \left|OQ\right|=\frac{12}{4+\sqrt{3}}$