Question

If the normal to the parabola $y^2=4ax$ at the point $(a{t}^2,2at)$ cuts the parabola again at $(a{T}^2,2aT)$, then the range of $T$ is

• A. $T\in (-2,2)$
• B. $T\in (-\infty ,-8)\cup (8,\infty )$
• C. $T\in (-2\sqrt{2},2\sqrt{2})$
• D. $T\in [-\infty ,-2\sqrt{2}]\cup [2\sqrt{2},\infty ]$

Answer 1

The correct option is D $T\in [-\infty ,-2\sqrt{2}]\cup [2\sqrt{2},\infty ]$

Given parabola is $y^2=4ax$
Equation of normal at $(a{t}^2,2at)$ is
$y+tx=2at+a{t}^3\cdots \left(1\right)$

This cuts again at $(a{T}^2,2aT)$, so
$2aT+ta{T}^2=2at+a{t}^3\Rightarrow 2T+t{T}^2=2t+t^3(\because a\ne 0)\Rightarrow t(t^2-T^2)+2(t-T)=0\Rightarrow (t-T)\left(t\right(t+T)+2)=0\Rightarrow t^2+Tt+2=0(\because t\ne T)$
As $t\in R$, so
$D\ge 0\Rightarrow T^2-8\ge 0\therefore T\in [-\infty ,-2\sqrt{2}]\cup [2\sqrt{2},\infty ]$


Alternate solution :
If normal at $(a{t}^2,2at)$ again cuts the parabola at $(a{T}^2,2aT)$, then
$T=-t-\frac{2}{t}{t}^2+Tt+2=0$
As $t\in R$, so
$D\ge 0\Rightarrow T^2-8\ge 0\therefore T\in [-\infty ,-2\sqrt{2}]\cup [2\sqrt{2},\infty ]$