Question

The tangent and normal at the point $P(a{t}^2,2at)$ to the parabola $y^2=4ax$ meet the x-axis in $T$ and $G$ respectively, then the angle at which the tangent at $P$ to the parabola is inclined to the tangent at $P$ to the circle through $P,T,G$ is

• A. ${\tan }^{-1}\left(t^2\right)$
• B. ${\cot }^{-1}\left(t^2\right)$
• C. ${\tan }^{-1}\left(t\right)$
• D. ${\cot }^{-1}\left(t\right)$

Answer 1

Answer: (C)

${\tan }^{-1}\left(t\right)$

Given equation of parabola is
$y^2=4ax$
Focus is at $(a,0)$
Equation of tangent to the parabola at $P(a{t}^2,2at)$ is
$ty=x+a{t}^2$ ....... (i)
Since, the tangent meets x-axis i.e. $y=0$
$\Rightarrow x=-a{t}^2$
So, coordinates of point $T$ is $(-a{t}^2,0)$
Equation of normal at $P(a{t}^2,2at)$ is
$y=-tx+2at+a{t}^3$ ... (ii)
Since, the normal meets x-axis i.e. $y=0$
$\Rightarrow x=2a+a{t}^2$
So, coordinates of $G$ are $(2a+a{t}^2,0)$
Since, the circle passes through $P,T,G$ so, its center is same as the focus.
So, center of the circle is $(a,0)$
Since $\angle TPG={90}^o$.
$({90}^o-\theta )$ is the angle between $PT$ and $OP$
Slope of $PT=\frac{2at}{2a{t}^2}=\frac{1}{t}$
Slope of $OP=\frac{2at}{a(t^2-1)}=\frac{2t}{t^2-1}$
$\therefore \tan ({90}^o-\theta )=\left|\frac{\frac{1}{t}-\frac{2t}{t^2-1}}{1+\frac{1}{t}\left(\frac{2t}{t^2-1}\right)}\right|=\frac{1}{t}$
$\therefore \cot \theta =\frac{1}{t}\Rightarrow \tan \theta =t$
$\Rightarrow \theta ={\tan }^{-1}\left(t\right)$