If the equal sides $PQ$ and $PR$ (each equal to 2) of a right angled isosceles $\triangle PQR$ be produced to $A$ and $B$ so that $OA.RB = PR^2$ then the line $AB$ passes through a fixed point which also satisfies the line $ax + by - 6 = 0$ then $a + b$ is ____.
Step-by-Step Solution
Key Concept: The constraint $OA \cdot RB = 4$ creates a hyperbolic relationship between the intercepts, forcing the line $AB$ to satisfy a parametric family of equations.
Let $QA = a$. Since $OA \cdot RB = PR^2 = 4$, we have $RB = \frac{4}{a}$. Point $A$ is at $(2+a, 0)$ and $B$ is at $(0, 2+\frac{4}{a})$. The equation of line $AB$ is $y = \frac{-(4+2a)}{a(2+a)}(x - 2 - a)$, which simplifies to $-a(2+a)y = (4+2a)(x-2-a)$.
Correct Answer: 3