$${{dx} \over {d\theta }} = - a\sin \theta $$ and $${{dy} \over {d\theta }} = a\cos \theta $$

$$\therefore$$ $${{dy} \over {dx}} = - \cot \theta .$$

$$\therefore$$ The slope of the normal at $$\theta $$ = $$ - {1 \over { - \cot \theta }}$$$$= \tan \theta $$

$$\therefore$$ The equation of the normal at $$\theta $$ is

$$y - a\sin \theta = \tan \theta \left( {x - a - a\cos \theta } \right)$$

$$ \Rightarrow $$ $$y - a\sin \theta =$$ $${{\sin \theta } \over {\cos \theta }}$$ $$\left( {x - a - a\cos \theta } \right)$$

$$ \Rightarrow y\cos \theta - a\sin \theta \cos \theta $$

$$\,\,\,\,\,\,\,\,\,\,\,$$ $$ = x\sin \theta - a\sin \theta - a\sin \theta \cos \theta $$

$$ \Rightarrow x\sin \theta - y\cos \theta = a\sin \theta $$

$$ \Rightarrow y = \left( {x - a} \right)\tan \theta $$

which always passes through $$(a, 0)$$

Let us define a function

$$f\left( x \right) = {{ax{}^3} \over 3} + {{b{x^2}} \over 2} + cx$$

Being polynomial, it is continuous and differentiable, also,

$$f\left( 0 \right) = 0\,$$ and $$\,\,f\left( 1 \right) = {a \over 3} + {b \over 2} + c$$

$$ \Rightarrow f\left( 1 \right) = {{2a + 3b + 6c} \over 6} = 0$$ (given)

$$\therefore$$ $$f\left( 0 \right) = f\left( 1 \right)$$

$$\therefore$$ $$f(x)$$ satisfies all conditions of Rolle's theorem

therefore $$f'\left( x \right) = 0$$ has a root in $$\left( {0,1} \right)$$

i.e. $$a{x^2} + bx + c = 0$$ has at lease one root in $$(0, 1)$$

$$f''\left( x \right) = 6\left( {x - 1} \right).$$ Inegrating,

we get $$f'\left( x \right) = 3{x^2} - 6x + c$$

Slope at $$\left( {2,1} \right) = f'\left( 2 \right) = c = 3$$

$$\left[ {\,\,} \right.$$ As slope of tangent at $$(2, 1)$$ is $$3$$ $$\left. {\,\,} \right]$$

$$\therefore$$ $$f'\left( x \right) = 3{x^2} - 6x + 3 = 3{\left( {x - 1} \right)^2}$$

Inegrating again, we get

$$f\left( x \right) = {\left( {x - 1} \right)^3} + D$$

The curve passes through $$(2,1)$$

$$ \Rightarrow 1 = {\left( {2 - 1} \right)^3} + D \Rightarrow D = 0$$

$$\therefore$$ $$f\left( x \right) = {\left( {x - 1} \right)^3}$$

$${y^2} = 18x \Rightarrow 2y{{dy} \over {dx}} = 18 \Rightarrow {{dy} \over {dx}} = {9 \over y}$$

Given $${{dy} \over {dx}} = 2 \Rightarrow {9 \over 2} = 2 \Rightarrow y = {9 \over 2}$$

Puting in $${y^2} = 18x \Rightarrow x = {9 \over 8}$$

$$\therefore$$ Required point is $$\left( {{9 \over 8},{9 \over 2}} \right)$$