Question

The tangent and normal at $P(t)$, 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)

Equation of tangent and normal at $P\left(a{t}^2,2at\right)$ on $y^2=4ax$ are
$ty=x+a{t}^2$... (1)
$y+tx=2at+a{t}^3$... (2)
So $T\left(-a{t}^2,0\right)\&G\left(2a+a{t}^2,0\right)$
equation of circle passing $P,T$ & $G$ is
$\left(x+a{t}^2\right)\left(x-\left(2a+a{t}^2\right)\right)+(y-0)(y-0)=0$
$x^2+y^2-2ax-a{t}^2\left(2a+a{t}^2\right)=0$
equation of tangent on the above circle at
$P\left(a{t}^2,2at\right)$ is $a{t}^{2}x+2aty-a\left(x+a{t}^2\right)-a{t}^2\left(2a+a{t}^2\right)=0$
slope of line which is tangent to circle at $P$
$m_1=\frac{a\left(1-t^2\right)}{2at}=\frac{1-t^2}{2t}$
slope of tangent at $P,m_2=\frac{1}{t}$
$\therefore \tan \theta =\frac{\frac{1-t^2}{2t}-\frac{1}{t}}{1+\frac{\left(1-t^2\right)}{2t^2}}\Rightarrow \tan \theta =t$
$\Rightarrow \theta ={\tan }^{-1}t={\sin }^{-1}\frac{t}{\sqrt{1+t^2}}$