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(D) Since $P(b, c)$ lies on the parabola $y^2 = 2ax$,we have $c^2 = 2ab$. The equation of the tangent to the parabola $y^2 = 2ax$ at point $(b, c)$ is

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$146$

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(D) Since $P(b, c)$ lies on the parabola $y^2 = 2ax$,we have $c^2 = 2ab$. The equation of the tangent to the parabola $y^2 = 2ax$ at point $(b, c)$ is $yc = a(x + b)$. To find the intersection with the $x$-axis $(y = 0)$,we set $y = 0$ in the tangent equation,which gives $0 = a(x + b)$,so $x = -b$. The triangle is formed by the tangent,the line $x = b$,and the line $y = 0$. The vertices are $P(b, c)$,the intersection of the tangent with $x = b$ (which is $(b, c)$),and the intersection of the tangent with $y = 0$ (which is $(-b, 0)$). The third vertex is the intersection of $x = b$ and $y = 0$,which is $(b, 0)$. The base of the triangle along the $x$-axis is the distance between $x = -b$ and $x = b$,which is $2b$. The height is the $y$-coordinate of $P$,which is $c$. Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes (2b) 	imes c = bc = 16$. Since $b, c \in \mathbb{N}$,the possible pairs $(b, c)$ are $(1, 16), (2, 8), (4, 4), (8, 2), (16, 1)$. From $c^2 = 2ab$,we have $a = \frac{c^2}{2b}$. For $(b, c) = (1, 16)$,$a = \frac{16^2}{2(1)} = \frac{256}{2} = 128$. For $(b, c) = (2, 8)$,$a = \frac{8^2}{2(2)} = \frac{64}{4} = 16$. For $(b, c) = (4, 4)$,$a = \frac{4^2}{2(4)} = \frac{16}{8} = 2$. For $(b, c) = (8, 2)$,$a = \frac{2^2}{2(8)} = \frac{4}{16} = 0.25$ (not a natural number). For $(b, c) = (16, 1)$,$a = \frac{1^2}{2(16)} = \frac{1}{32}$ (not a natural number). Thus,$S = \{128, 16, 2\}$. The sum is $128 + 16 + 2 = 146$.

Let $S$ be the set of all $a \in \mathbb{N}$ such that the area of the triangle formed by the tangent at the point $P(b, c)$,where $b, c \in \mathbb{N}$,on the parabola $y^2 = 2ax$ and the lines $x = b$ and $y = 0$ is $16 \text{ unit}^2$. Then $\sum_{a \in S} a$ is equal to $..........$.

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