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 $|r|$ is
Step-by-Step Solution
Key Concept: Use the distance formula and the symmetry of curves about $y = x$ to establish relationships between parameters.
Let $A = (t, \frac{t}{2})$ and $B = (t, -t)$. Both curves are symmetric about the line $y = x$. Then $(AB)^2 = (t - t)^2 + (\frac{t}{2} - (-t))^2 = (\frac{3t}{2})^2 = \frac{9t^2}{4}$. So $2(t^2 + \frac{t^2}{4}) = 14 + 4t^2 + \frac{t^2}{9}$. From $(OA)^2 = t^2 + \frac{t^2}{4} = 9 \Rightarrow (2t)^2 = 9$, giving $|2t| = 3$, so $|2t| = 3 \Rightarrow t = 1.5$.
Correct Answer: 1.5