If the normals to the curve $y = x^2$ at the points $P, Q$ & $R$ passes through the point $(0, 3/2)$, find the radius of the circle circumscribing $\triangle PQR$.
Step-by-Step Solution
Key Concept: Find points where normals to the parabola y = x² pass through (0, 3/2) using the normal equation x + 2ty = t + 2t³, then determine if the circumcircle has a special property (right angle at a vertex implies circumradius = hypotenuse/2).
For the parabola $y = x^2$, the normal at point $(t, t^2)$ has equation $x + 2y = t + 2t^3$. When this normal passes through $(0, 3/2)$, we get $t^3 = t$, giving $t = 0, 1, -1$ and three points $P(0,0)$, $Q(1,1)$, $R(-1,1)$. Since $\angle P = 90°$, the circumradius is $R = \frac{QR}{2} = 1$.
Correct Answer: 1