If a chord, which is not a tangent, of the parabola $y^{2} = 16x$ has the equation $2x + y = p,$ and midpoint (h, k), then which of the following is(are) possible value(s) of p, h and $k ? (1) p = -2, h = 2, k = -4 (2) p = -1, h = 1, k = -3 (3) p = 2, h = 3, k = -4 (4) p = 5, h = 4, k = -3$

$\text{Given: parabola } y^2 = 16x$

$\text{chord } 2x + y = p \ \text{(not a tangent)}$

$\text{midpoint } (h, k)$

$\text{To find: possible values of } p, h, k$

$y^2 = 4ax \ \Rightarrow \ 4a = 16 \ \Rightarrow \ a = 4$

$\text{vertex } (0,0), \quad \text{focus } (4, 0)$

$P(t) = (4t^2, 8t)$

$\text{chord joins } t_1 \text{ and } t_2$

$\text{slope} = \dfrac{2}{t_1 + t_2} = -2 \ \Rightarrow \ t_1 + t_2 = -1$

$k = \dfrac{8t_1 + 8t_2}{2} = 4(t_1 + t_2) = -4$

$h = \dfrac{4t_1^2 + 4t_2^2}{2} = 2\big[(t_1+t_2)^2 - 2t_1 t_2\big]$

$h = 2 - 4t_1 t_2$

$y(t_1+t_2) = 2x + 8t_1 t_2 \ \Rightarrow \ 2x + y = -8t_1 t_2$

$p = -8t_1 t_2$

$p = -8t_1 t_2 = -2(2 - h) = 2h - 4$

$\text{not a tangent} \Rightarrow t_1 \ne t_2 \Rightarrow (t_1 - t_2)^2 > 0$

$1 - 4t_1 t_2 > 0 \Rightarrow 1 - (2 - h) > 0 \Rightarrow h > 1$

$k = -4, \quad p = 2h - 4, \quad h > 1$

hline (p,h,k) condition valid? hline (-2,2,-4) p ≠ 2h-4 × (-1,1,-3) k ≠ -4 × (2,3,-4) k=-4, p=2h-4 checkmark (5,4,-3) k ≠ -4 × hline

$\boxed{\text{Answer: option (3)} \quad p = 2,\ h = 3,\ k = -4}$