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

Let $A=\left(t, \frac{1}{ t }\right) \& B=\left(\frac{1}{t}, t\right)$ $\begin{Bmatrix} \because Both \, curves \, are \, symmetric \, \\ about \, line \, y=x \end{Bmatrix}$
then, $\left(A B\right)^{2}=\left(t - \frac{1}{t}\right)^{2}+\left(\frac{1}{t} - t\right)^{2}=14$
$\Rightarrow 2\left(t^{2} + \frac{1}{t^{2}}\right)=14+4\Rightarrow t^{2}+\frac{1}{t^{2}}=9$
Now, $\left(O A\right)^{2}=t^{2}+\frac{1}{t^{2}}=9\Rightarrow \left(2 r\right)^{2}=9$
$\Rightarrow \left|2 r\right|=3$ $\Rightarrow \left|r\right|=1.5$