Parabola Part 2

Concept:For a parabola $y^2 = 4ax$, the ends of a focal chord have parameters $t_1$ and $t_2$ satisfying $t_1 t_2 = -1$.Here $y^2 = 16x$ gives $a = 4$.Explanation:Given one end $A(16,16)$ corresponds to $t_1 = 2$ because $x = 4t_1^2 = 16$ and $y = 8t_1 = 16$.Using $t_1 t_2 = -1$, we get $t_2 = -\frac{1}{2}$.So the other end $B$ is $(1, -4)$.The focal chord $AB$ is divided internally by $P(\alpha, \beta)$ in the ratio $5:2$.Using section formula, there are two possible points depending on which segment ratio refers to.If $AP:PB = 5:2$, then$\alpha = \frac{2 \cdot 16 + 5 \cdot 1}{7} = \frac{37}{7}$, $\beta = \frac{2 \cdot 16 + 5 \cdot (-4)}{7} = \frac{12}{7}$.If $AP:PB = 2:5$, then$\alpha = \frac{5 \cdot 16 + 2 \cdot 1}{7} = \frac{82}{7}$, $\beta = \frac{5 \cdot 16 + 2 \cdot (-4)}{7} = \frac{72}{7}$.Thus $\alpha + \beta$ can be $\frac{37}{7} + \frac{12}{7} = 7$ or $\frac{82}{7} + \frac{72}{7} = 22$.The minimum value is $7$.Answer:$7$ (Option B).