The normal at the point $ \left( a,2a \right) $ on $ {{y}^{2}}=4ax $ meets the curve again at $ \left( a{{t}^{2}},2at \right) $ . Then the value of its parameter is equal to
(a) 1
(b) 3
(c) –1
(d) –3

Hint: We start solving the problem by drawing the figure representing the given information. We then find the slope of the tangent at the point $ \left( a,2a \right) $ using the fact that the slope of the tangent to a curve at a point $ \left( {{x}_{1}},{{y}_{1}} \right) $ is defined as $ {{\left. \dfrac{dy}{dx} \right|}_{\left( {{x}_{1}},{{y}_{1}} \right)}} $ . We then find the slope of the normal using the fact that the product of slopes of tangent and normal is ‘–1’ as they both are perpendicular to each other. Since the normal is passing through the point $ \left( a,2a \right) $ and $ \left( a{{t}^{2}},2at \right) $ , we make use of the fact that the slope of the line passing through the points $ \left( {{x}_{1}},{{y}_{1}} \right) $ and $ \left( {{x}_{2}},{{y}_{2}} \right) $ is $ \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} $ to get the required answer.

Complete step by step answer:
According to the problem, we have given that the normal at the point $ \left( a,2a \right) $ on $ {{y}^{2}}=4ax $ meets the curve again at $ \left( a{{t}^{2}},2at \right) $ . We need to find the value of the parameter.
Let us draw the figure to represent the given information.



Let us first find the slope of the tangent at the point $ \left( a,2a \right) $ .
We know that the slope of the tangent to a curve at a point $ \left( {{x}_{1}},{{y}_{1}} \right) $ is defined as $ {{\left. \dfrac{dy}{dx} \right|}_{\left( {{x}_{1}},{{y}_{1}} \right)}} $ .
Let us differentiate both sides of $ {{y}^{2}}=4ax $ w.r.t x.
So, we get $ \dfrac{d}{dx}\left( {{y}^{2}} \right)=\dfrac{d}{dx}\left( 4ax \right) $ .
 $ \Rightarrow 2y\dfrac{dy}{dx}=4a $ .
 $ \Rightarrow \dfrac{dy}{dx}=\dfrac{4a}{2y} $ .
Now, the slope of the tangent to the curve $ {{y}^{2}}=4ax $ at $ \left( a,2a \right) $ is $ m={{\left. \dfrac{dy}{dx} \right|}_{\left( a,2a \right)}}=\dfrac{4a}{2\left( 2a \right)} $ .
 $ \Rightarrow m=\dfrac{4a}{4a}=1 $ .
We know that the product of slopes of tangent and normal is ‘–1’ as they both are perpendicular to each other. Let us assume the slope of tangent be $ {{m}_{1}} $ .
So, we get $ {{m}_{1}}\times 1=-1\Leftrightarrow {{m}_{1}}=-1 $ .
According to the problem, we are given that this normal also passes through the point $ \left( a{{t}^{2}},2at \right) $ . This means that the slope of the line passing through the points $ \left( a,2a \right) $ and $ \left( a{{t}^{2}},2at \right) $ is –1.
We know that the slope of the line passing through the points $ \left( {{x}_{1}},{{y}_{1}} \right) $ and $ \left( {{x}_{2}},{{y}_{2}} \right) $ is $ \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} $ .
So, we get $ \dfrac{a{{t}^{2}}-a}{2at-2a}=-1 $ .
 $ \Rightarrow \dfrac{a\left( {{t}^{2}}-1 \right)}{2a\left( t-1 \right)}=-1 $ .
 $ \Rightarrow \dfrac{t+1}{2}=-1 $ .
 $ \Rightarrow t+1=-2 $ .
 $ \Rightarrow t=-3 $ .
So, we have found the value of the parameter as –3.
 $ \, therefore, $ The correct option for the given problem is (d).
Note:
We can see that the problem has a huge amount of calculation so, we need to perform each step carefully in order to avoid confusion and mistakes. We can also solve this problem by finding the equation of the normal and then finding the intersection points of normal and parabola. We can also find the equation of the tangent at the point $ \left( a{{t}^{2}},2at \right) $ . Similarly, we can expect problems to find the angle made by the points $ \left( a,2a \right) $ , $ \left( a{{t}^{2}},2at \right) $ at the vertex of the parabola.