Q114 / 200 · Manipal (MET) 2010 · Mathematics
If the normal to the parabola $y^2=4 a x$ at the point $\left(a t^2, 2 a t\right)$ cuts the parabola again at $\left(a T^2, 2 a T\right)$, then
- A$-2 \leq T \leq 2$
- B$T \in(-\infty,-8) \cup(8, \infty)$
- C$T^2 \lt 8$
- Correct answerD$T^2 \geq 8$
Solution
Equation of the normal of the parabola $y^2=4 a x$ at the point $\left(a t^2, 2 a t\right)$ is $y+t x=2 a t+a t^3...(i)$
$\because$ Eq. (i) cuts the parabola again at $\left(a T^2, 2 a T\right)$.
Then, $2 a T+t a T^2=2 a t+a t^3$
$\Rightarrow \quad 2 a(T-t)=-a t\left(T^2-t^2\right)$
$\Rightarrow \quad 2=-t(T+t) \quad(\because t \neq T)$
$\Rightarrow \quad t^2+t T+2=0$
$\because t$ is real, $\therefore T^2-4 \cdot 1 \cdot 2 \geq 0$
$\Rightarrow \quad T^2 \geq 8$
Hence, the correct answer is option: 'D'.