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(a) \[{{\cot }^{-1}}t\]

(b) \[{{\cot }^{-1}}{{t}^{2}}\]

(c) \[{{\tan }^{-1}}t\]

(d) \[{{\tan }^{-1}}{{t}^{2}}\]

Answer
Verified

We know that equation of parabola \[{{y}^{2}}=4ax\]. Let P be the point where the tangent meets the parabola

The tangent meets the x axis at point T and the normal at point G.

Let point P be \[\left( a{{t}^{2}},2at \right)\approx \left( x,y \right)\].

The slope of the tangent can be found by using,

\[{{m}_{T}}=\dfrac{y-{{y}_{1}}}{x-{{x}_{1}}}=\dfrac{2at-0}{a{{t}^{2}}-0}=\dfrac{2at}{a{{t}^{2}}}=\dfrac{\Delta y}{\Delta x}\]

\[{{m}_{T}}=\dfrac{2a}{2at}=\dfrac{1}{t}\]

The slope of the normal will be \[=-\left( \dfrac{1}{{{m}_{T}}} \right)=-\left( \dfrac{1}{\dfrac{1}{t}} \right)=-t\]

\[\therefore {{M}_{n}}=-t\]

We know the equation of tangent,

\[\Rightarrow y=mx+c....\left( i \right)\]

We got the slope m as \[\left( \dfrac{1}{t} \right)\] for the tangent.

Therefore, equation (ii) of the tangent changes to

\[\Rightarrow y=\dfrac{x}{t}+c....\left( ii \right)\]

We know the value of \[y=2at\] and \[x=a{{t}^{2}}\].

Substituting these values in equation (ii) along with slope \[m=\dfrac{1}{t}\].

We get,

\[2at=\dfrac{a{{t}^{2}}}{t}+c\]

By simplifying it, we get

\[2at=at+c\]

\[\Rightarrow c=2at-at\]

\[\therefore c=at\]

Now, put the value of c in equation (ii).

\[\Rightarrow y=\dfrac{x}{t}+at....\left( iii \right)\]

Now, put the value of y = 0, to get the coordinates of point T.

\[0=\dfrac{x}{t}+at\]

\[\Rightarrow x=\left( -at \right)t=-a{{t}^{2}}\]

Therefore, Coordinates of T are \[\left( -a{{t}^{2}},0 \right)\].

We got the slope of Normal, \[{{m}_{N}}=-t\].

By putting the value of slope in \[y=mx+c\], we get

\[y=-tx+c....\left( iv \right)\]

Put \[\left( x,y \right)=\left( a{{t}^{2}},2at \right)\] in the above equation.

\[2at=-t\left( a{{t}^{2}} \right)+c\]

\[\Rightarrow 2at+a{{t}^{3}}=c\]

\[\therefore c=at\left( 2+a{{t}^{2}} \right)\]

Now, put the value of c in equation (iv)

\[y=-tx+at\left( 2+a{{t}^{2}} \right)\]

\[\Rightarrow y=-tx+2at+a{{t}^{3}}\]

Put the value of y = 0, to get the coordinates of point G.

\[0=-tx+2at+a{{t}^{3}}\]

\[\Rightarrow tx=2at+a{{t}^{3}}\]

Dividing throughout with ‘t’ on both RHS and LHS

\[\dfrac{tx}{x}=\dfrac{2at+a{{t}^{3}}}{t}\]

\[\Rightarrow x=2a+a{{t}^{2}}\]

Therefore, we get the coordinates of G as \[\left( 2a+a{{t}^{2}},0 \right)\].

Now, let us take the focus of parabola as F with coordinates (a, 0).

Now, calculate the distance of FT by using the formula

\[\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( \dfrac{1}{2}-\dfrac{1}{1} \right)}^{2}}}\]

Therefore, distance of |FT| where \[F\left( a,0 \right)\] and \[T\left( -a{{t}^{2}},0 \right)\].

\[\left| FT \right|=\sqrt{{{\left[ a-\left( -a{{t}^{2}} \right) \right]}^{2}}+\left[ 0-0 \right]}=\sqrt{{{\left( a+a{{t}^{2}} \right)}^{2}}}=a+a{{t}^{2}}\]

Similarly distance between |FG| where \[F\left( a,0 \right)\] and \[G\left( 2a+a{{t}^{2}},0 \right)\].

\[\left| FG \right|=\sqrt{\left( 2a+a{{t}^{2}}-{{a}^{2}} \right)+\left( 0-0 \right)}=\sqrt{{{\left( a+a{{t}^{2}} \right)}^{2}}}=a+a{{t}^{2}}\]

Similarly, if we find the distance of |PF|

Therefore, the directrix of parabola will be x + a = 0 from x axis and then \[\left( x-a{{t}^{2}} \right)\] to point p on the other side of x – axis.

Therefore from the figure, the distance \[\left| PF \right|=a+a{{t}^{2}}\text{ }\]

Now, we can create the circle connecting the points T, P and G.

So focus F becomes the center of the circle.

We know that the equation of the circle,

\[{{\left( x-g \right)}^{2}}+{{\left( y-f \right)}^{2}}={{r}^{2}}....\left( v \right)\]

where (g, f) is the center of the circle.

Substituting, (g, f) as (a, 0) in equation (v), we get

\[{{\left( x-a \right)}^{2}}+{{\left( y-0 \right)}^{2}}={{\left( a+a{{t}^{2}} \right)}^{2}}\]

where radius, \[r=a+a{{t}^{2}}\]

Now, differentiate \[{{\left( x-a \right)}^{2}}+{{y}^{2}}={{\left( a+a{{t}^{2}} \right)}^{2}}\], we get

\[2\left( x-a \right)+2y.\dfrac{dy}{dx}=0\]

By deriving the value of \[\dfrac{dy}{dx}=\dfrac{-2\left( x-a \right)}{2y}\]

\[\therefore \dfrac{dy}{dx}=\dfrac{-\left( x-a \right)}{y}=\dfrac{a-x}{y}\]

Substitute the value of x and y in the above equation,

i.e. \[\left( x,y \right)=\left( a{{t}^{2}},2at \right)\]

\[\therefore \dfrac{dy}{dx}=\dfrac{a-a{{t}^{2}}}{2at}=\dfrac{a\left( 1-{{t}^{2}} \right)}{2at}=\dfrac{1-{{t}^{2}}}{2t}={{m}_{2}}\]

i.e the slope of the circle.

Now, let us draw a tangent on the circle whose slope is equal to the tangent through P. The angle between the new tangent and circle will be \[\theta \]. Here focus is taken as the center of the circle formed by points P, G and T. The focus is at the intersection of the axis and the circle. Therefore, it becomes the center of the circle.

\[\therefore \tan \theta =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|\]

This is to find the angle made by the tangent and the circle.

We know the slope, \[{{m}_{1}}=\dfrac{1}{t}\], i.e. the slope of the tangent.

The slope of the circle, \[{{m}_{2}}=\dfrac{1-{{t}^{2}}}{2t}\]

\[\therefore \tan \theta =\left| \dfrac{\dfrac{1}{t}-\left( \dfrac{1-{{t}^{2}}}{2t} \right)}{1+\left( \dfrac{1}{t} \right)\left( \dfrac{1-{{t}^{2}}}{2t} \right)} \right|=\left| \dfrac{\dfrac{2-1+{{t}^{2}}}{2t}}{1+\dfrac{1-{{t}^{2}}}{2{{t}^{2}}}} \right|\]

\[=\left| \dfrac{1+\dfrac{{{t}^{2}}}{2t}}{\dfrac{2{{t}^{2}}+1-{{t}^{2}}}{2{{t}^{2}}}} \right|=\left| \dfrac{1+{{t}^{2}}}{1+\dfrac{{{t}^{2}}}{t}} \right|=\left| \dfrac{t\left( 1+{{t}^{2}} \right)}{1+{{t}^{2}}} \right|\]

\[\therefore \tan \theta =t\]

\[\therefore \theta ={{\tan }^{-1}}t\]