Question

Let $x^2+y^2=4r^2$ and $xy=1$ intersects at $A$ and $B$ in first quadrant. If $AB=\sqrt{14}$ units, then the value of $\left|r\right|$ is

Answer 1

Answer: 1.5

Let $A=\left(t,\frac{1}{t}\right)\&B=\left(\frac{1}{t},t\right)$ $\left\{\begin{array}{c}\because Bothcurvesaresymmetric\\ aboutliney=x\end{array}\right\}$
then, ${\left(AB\right)}^2={\left(t-\frac{1}{t}\right)}^2+{\left(\frac{1}{t}-t\right)}^2=14$
$\Rightarrow 2\left(t^2+\frac{1}{t^2}\right)=14+4\Rightarrow t^2+\frac{1}{t^2}=9$
Now, ${\left(OA\right)}^2=t^2+\frac{1}{t^2}=9\Rightarrow {\left(2r\right)}^2=9$
$\Rightarrow \left|2r\right|=3$ $\Rightarrow \left|r\right|=1.5$