Question

The tangent and normal at $P\left(t\right)$, for all real positive $t$, to the parabola $y^2=4ax$ meet the axis of the parabola 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 the points $P,T$ and $G$ is -

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

Answer 1

Answer: (C), (D)

${\sin }^{-1}\left(\frac{t}{\sqrt{1+t^2}}\right)$
${\tan }^{-1}t$

Equation of tangent and normal at $P(a{t}^2,2at)$ on $y^2=4ax$ are
$ty=x+a{t}^2$ .....(1)
$y+tx=2at+a{t}^3$ ...(2)
So, $T(-a{t}^2,0)$ & $G(2a+a{t}^2,0)$

Equation of circle passing through $P,T$ and $G$ is
$(x+a{t}^2)(x-(2a+a{t}^2\left)\right)+(y-0)(y-0)=0$
$x^2+y^2-2ax-a{t}^2(2a+a{t}^2)=0$
Equation of tangent on the above circle at $P(a{t}^2,2at)$ is $a{t}^{2}x+2aty-a(x+a{t}^2)-a{t}^2(2a+a{t}^2)=0$
Slope of line which is tangent to circle at $P$
$m_1=\frac{a(1-t^2)}{2at}=\frac{1-t^2}{2t}$
Slope of tangent to parabola at $P$,

$m_2=\frac{1}{t}$
$\therefore \tan \theta =\frac{\frac{1-t^2}{2t}-\frac{1}{t}}{1+\frac{(1-t^2)}{2t^2}}$

$\Rightarrow \tan \theta =t$
$\Rightarrow \theta ={\tan }^{-1}t={\sin }^{-1}\frac{t}{\sqrt{1+t^2}}$