<p>Normals of parabola \(y^2 = 4x\) at P and Q meet at \(R(x_2, 0)\). If \(x_2 = 4\) and area of circle circumscribing \(\triangle PQR\) is \(k\pi\), then \(k =\)</p>
Step-by-Step Solution
Key Concept: Find the circumradius of the triangle formed by two points on parabola and normal meeting point.
<p>For \(x_2 = 4\): \(t_1^2 + t_2^2 + 2 = 4 \Rightarrow t_1^2 + t_2^2 = 2\)</p><p>Let \(t_1t_2 = c\). Then \((t_1 + t_2)^2 = 2 + 2c\) and \((t_1 - t_2)^2 = 2 - 2c\)</p><p>Points: P: \((t_1^2, 2t_1)\), Q: \((t_2^2, 2t_2)\), R: \((4, 0)\)</p><p>Using circumradius formula \(R = \frac{abc}{4K}\) where a, b, c are sides and K is area:</p><p>After calculation, Area of circumcircle = \(9\pi\), so \(k = 9\)</p>
Correct Answer: s