If $f (x)$ is a polynomial with positive leading coefficient satisfying $f (0) = 0$ $f (f (x)) = x \int_{0}^{x} f (t) d t \forall x \in R,$ then locus of point of intersection of two perpendicular tangents can be drawn to the curve $y = f (x)$ can be given by $a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + \sqrt{3} = 0$ , then the value of $a^{2} - \sqrt{5} h^{3} + \sqrt{2} b^{4} + \sqrt{3} g^{5} + f =$

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Detailed Solution

Let the degree of $f (x)$ be $' n'$
So, we get $n^{2} = 1 + n + 1 \Rightarrow n = 2$
Let $f (x) = a x^{2} + b x$ as $f (0) = 0$
Given relation $f (f (x)) = x \int_{0}^{x} f (t) d t$

$\Rightarrow a {(a x^{2} + b x)}^{2} + b (a x^{2} + b x) = x (\frac{a x^{3}}{3} + \frac{b x^{2}}{2})$

$\Rightarrow a^{3} = \frac{a}{3} & 2 a^{2} b = \frac{b}{2} & a b^{2} + a b = 0 & b^{2} = 0$

$\Rightarrow b = 0 & a = \frac{1}{\sqrt{3}}$

So, we get $f (x) = \frac{x^{2}}{\sqrt{3}}$
So, the directrix will be $y = - \frac{4}{\sqrt{3}}$

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