Parabola $y^2 = 4x$ and the circle having its centre at $(6, 5)$ intersect at right angle. Possible point of intersection of these curves can be :
Step-by-Step Solution
Key Concept: Two curves intersect at right angles when their tangents at the intersection point are perpendicular. For parabola y² = 4x, the tangent at point (t², 2t) has slope 1/t, and this tangent must pass through the circle's center (6, 5) for orthogonal intersection, yielding the condition t² - 5t + 6 = 0.
A point on the parabola $y^2 = 4x$ is $(t^2, 2t)$. The tangent at this point is $y = x + t^2$, which must pass through $(6, 5)$. This gives $t^2 - 5t + 6 = 0$, so $t = 2, 3$. The possible points are $(4, 4)$, $(4, -4)$, and $(9, 6)$.
Correct Answer: 1,3