Let $x^2+y^2=r^2$ and $xy=1$ intersect at $A$ and $B$ in the first quadrant. If $AB=\sqrt{14}$ units, then the square of the distance of $AB$ from the origin is equal to

Solution

Let, $A=\left(t,\frac{1}{t}\right)$ and $B=\left(\frac{1}{t},t\right)$ { $\because$ both curves are symmetric about line $y=x$ }
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, ${\left(OA\right)}^2=t^2+\frac{1}{t^2}=9\Rightarrow r^2=9$
$\Rightarrow r=3$
$\Rightarrow O{F}^2=9-\frac{14}{4}=5.5$

Hence, the correct answer is 5.5.