Let S be the set of all a ∈ N such that the area of the triangle...

Given,

$S$ be the set of all $a \in N$ such that the area of the triangle formed by the tangent at the point $P\left(b, c\right), b, c \in N$, on the parabola ${y}^{2} = 2ax$

And the lines $x = b, y = 0$ is $16{unit}^{2}$

Now plotting the diagram we get,

Tangent at $$P\left(b, c\right)$ will be $ax - cy + ab = 0$ and it meets $x$-axis at \left(- b, 0\right)$</p><p>Now area of triangle$\left(PAB\right) = \dfrac{1}{2}$\left(2b\right)\left(c\right) = 16 ⇒ bc = 16$

So, \left(b, c\right) = \left(1, 16\right), \left(2, 8\right), \left(4, 4\right), \left(8, 2\right), \left(16, 1\right)$as both belongs to natural number,</p><p>Now$\left(b, c\right)$lie on parabola hence satisfy it</p><p>${c}^{2} = 2ab ⇒ {c}^{2} = 2a\left(\dfrac{16}{c}\right) ⇒ a = \dfrac{{c}^{3}}{32}$</p><p>Now when possible value of$c$is$c = 4, 8, 16 ⇒ a = 2, 16, 128$</p><p>Hence,$\mathop{∑}_{a \in S}a = 2 + 16 + 128 = 146$.