ParabolaHardBloom L3

Question

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

Options

A.$22$
B.$7$
C.$5$
D.$16$

Solution

**Given:** Parabola $y^2 = 16x$, so $4a = 16 \Rightarrow a = 4$, focus $S = (4, 0)$. Parametric form: $(4t^2,\, 8t)$. For point $A(16, 16)$: $8t = 16 \Rightarrow t = 2$ ✓ For a focal chord, if one end has parameter $t_1 = 2$, the other end has parameter: $$t_2 = -\frac{1}{t_1} = -\frac{1}{2}$$ So $B = \left(4 \cdot \tfrac{1}{4},\; 8 \cdot \left(-\tfrac{1}{2}\right)\right) = (1, -4)$. **Case 1:** $P$ divides $AB$ internally in ratio $5:2$ $$\alpha = \frac{5(1) + 2(16)}{7} = \frac{5 + 32}{7} = \frac{37}{7}$$ $$\beta = \frac{5(-4) + 2(16)}{7} = \frac{-20 + 32}{7} = \frac{12}{7}$$ $$\alpha + \beta = \frac{37 + 12}{7} = \frac{49}{7} = 7$$ **Case 2:** $P$ divides $AB$ internally in ratio $2:5$ $$\alpha = \frac{2(1) + 5(16)}{7} = \frac{2 + 80}{7} = \frac{82}{7}$$ $$\beta = \frac{2(-4) + 5(16)}{7} = \frac{-8 + 80}{7} = \frac{72}{7}$$ $$\alpha + \beta = \frac{82 + 72}{7} = \frac{154}{7} = 22$$ Minimum value of $\alpha + \beta = \min(7, 22) = 7$. **Answer: (B)**

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