The normal at four points $A, B, C$ and $D$ on the rectangular hyperbola $xy = c^2$ meet in $P(h,k)$ and $PA^2 + PB^2 + PC^2 + PD^2 = n\left(h^2 + k^2\right)$, where $n$ is equal to
Step-by-Step Solution
Key Concept: Expand the sum of squared distances and use the given constraints on the sum of coordinates and sum of squared coordinates to simplify.
For four points $A, B, C, D$ with coordinates $(x_i, y_i)$ for $i = 1,2,3,4$, and with $\sum x_i = h\sum 1, \sum y_i = k\sum 1, \sum x_i^2 = h^2, \sum y_i^2 = k^2$, the sum $PA^2 + PB^2 + PC^2 + PD^2 = \sum[(h-x_i)^2 + (k-y_i)^2] = 4(h^2+k^2) - 2h\sum x_i - 2k\sum y_i + \sum x_i^2 + \sum y_i^2 = 3(h^2+k^2)$.
<div class="key-concept"><strong>Key Concept:</strong> Expand the sum of squared distances and use the given constraints on the sum of coordinates and sum of squared coordinates to simplify.</div>
<div class="trap-box"><strong>Trap:</strong> Students often make algebraic errors when expanding multiple squared terms or fail to carefully substitute the constraint conditions.</div>
Correct Answer: 3