Question

Let $L$ be an end of the latus rectum of $y^2=4x$. If the normal at $L$ meets the curve again at $M$ and the normal at $M$ meets the curve again at $N,$ then area of $△LMN$ (in sq. units) is

• A. $\frac{1280}{9}$
• B. $\frac{640}{9}$
• C. $\frac{320}{9}$
• D. $\frac{160}{9}$

Answer 1

The correct option is B $\frac{640}{9}$

Parabola is $y^2=4x$, $a=1$
Point $L$ is $(a,2a)\equiv (a{t}^2,2at)$
$L\equiv (1,2),t=1$
Normal at $L\left(t\right)$ passes through $M\left(t_1\right)$
$\therefore t_1=-t-\frac{2}{t}$
$t=1$
$\Rightarrow t_1=-3$
Point $M$ is $(t_1^2,2t_1)\equiv (9,-6)$
So another point is $M(9,-6)$

and normal at $M$ passes through $N\left(t_2\right)$
$t_2=-t_1-\frac{2}{t_1}=\frac{11}{3}$
Point is $N\equiv (t_2^2,2t_2)$
$\Rightarrow N\equiv (\frac{121}{9},\frac{22}{3})$

Now area of $△LMN$ is
$\Delta =\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$
$\Delta =\frac{640}{9}$ sq. units