The tangents at a point $P$ to the rectangular hyperbola $xy = 1$ meets the lines $x - y = 0$ and $x + y = 0$ at $Q$ and $R$ respectively and $\Delta_1$ is the area of the triangle $OQR$, where $O$ is the origin. The normal at $P$ meets $X$-axis at $M$ and the $Y$-axis at $N$ and $\Delta_2$ is the area of the triangle $OMN$, then the value of $\Delta_1^2\Delta_2$ is \ldots.
Step-by-Step Solution
Key Concept: Use parametric coordinates on the hyperbola to find tangent and normal lines, then calculate areas of triangles formed by intersections with coordinate axes.
The point $P$ has coordinates $\left(t, \frac{1}{t}\right)$. The tangent at $P$ is $x + t^2y = 2t$, which meets $y = x$ and $y = -x$ at points $Q\left(\frac{2t}{1+t^2}, \frac{2t}{1+t^2}\right)$ and $R\left(\frac{2t}{1-t^2}, -\frac{2t}{1-t^2}\right)$. The area of triangle $OQR$ is $A_1 = \frac{4t^2}{|1-t^4|}$. The normal at $P$ is $t^3x - ty = t^4 - 1$, meeting the axes at $M\left(\frac{t^4-1}{t^3}, 0\right)$ and $N\left(0, \frac{1-t^4}{t}\right)$, giving area $A_2 = \frac{(t^4-1)^2}{2t^4}$. Setting $A_1 = A_2$ yields $A_1^2A_2 = 8$.
Correct Answer: 8