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

• A. $-2\le T\le 2$
• B. $T\in (-\infty ,-8)\cup (8,\infty )$
• C. $T^2<8$
• D. $T^2\ge 8$

Answer 1

The correct option is D $T^2\ge 8$

Eqution of the normal of the parabola $y^2=4ax$ at the point $(a{t}^2,2at)$ is
$y+tx=2at+a{t}^3...\left(i\right)$
$\because$ equation $(i$) cuts the parabola again at $(a{T}^2,2aT)$.
Then, $2aT+ta{T}^2=2at+a{t}^3$
$\Rightarrow 2a(T-t)=-at(T^2-t^2)$
$\Rightarrow t^2+tT+2=0$ $(\because t\ne T)$
$\because t$ is real, $\therefore T^2-4\cdot 1\cdot 2\ge 0\Rightarrow T^2\ge 8$