Question

The tangent and normal at $P ( t )$, for all real positive $t$, to the parabola $y ^2=4 ax$ 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 $P \left( at ^2, 2 at \right)$ on $y^2=4 a x$ are
$t y=x+a t^2$... (1)
$y+t x=2 a t+a t^3$... (2)
So $T\left(-a t^2, 0\right) \& G\left(2 a+a t^2, 0\right)$
equation of circle passing $P, T$ & $G$ is
$\left(x+a t^2\right)\left(x-\left(2 a+a t^2\right)\right)+(y-0)(y-0)=0 $
$x^2+y^2-2 a x-a t^2\left(2 a+a t^2\right)=0$
equation of tangent on the above circle at
$P\left(a t^2, 2 a t\right)$ is $a t^2 x+2 a t y-a\left(x+a t^2\right)-a t^2\left(2 a+a t^2\right)=0$
slope of line which is tangent to circle at $P$
$m _1=\frac{ a \left(1- t ^2\right)}{2 at }=\frac{1- t ^2}{2 t }$
slope of tangent at $P , m _2=\frac{1}{ t }$
$\therefore \tan \theta=\frac{\frac{1- t ^2}{2 t }-\frac{1}{ t }}{1+\frac{\left(1- t ^2\right)}{2 t ^2}} \Rightarrow \tan \theta= t$
$\Rightarrow \theta=\tan ^{-1} t =\sin ^{-1} \frac{ t }{\sqrt{1+ t ^2}}$