Let one end of a focal chord of the parabola $$y^{2}=16x$$ be (16,16). If $$P\left(\alpha,\beta\right)$$ divides this focal chord internally in the ratio 5 : 2, then the minimum value of $$\alpha+\beta$$ is equal to :

One end of a focal chord of $$y^2 = 16x$$ is $$(16, 16)$$. Here $$4a = 16$$, so $$a = 4$$.

Verify $$(16, 16)$$ lies on the parabola: $$16^2 = 256 = 16 \times 16$$ ✓.

Using parametric form: $$(at^2, 2at) = (4t^2, 8t)$$. For $$(16, 16)$$: $$8t = 16 \Rightarrow t = 2$$.

The other end of the focal chord has parameter $$t' = -1/t = -1/2$$.

Other end: $$(4 \times 1/4, 8 \times (-1/2)) = (1, -4)$$.

Point $$P(\alpha, \beta)$$ divides the focal chord from $$(16, 16)$$ to $$(1, -4)$$ in ratio $$5:2$$.

$$\alpha = \frac{5(1) + 2(16)}{7} = \frac{5 + 32}{7} = \frac{37}{7}$$

$$\beta = \frac{5(-4) + 2(16)}{7} = \frac{-20 + 32}{7} = \frac{12}{7}$$

$$\alpha + \beta = \frac{37 + 12}{7} = \frac{49}{7} = 7$$

The answer is Option 4: 7.