The length of latus rectum of the conic passing through the origin and having the property that normal at each point (x, y) intersects the x-axis at  $\left(\left(x+1\right),0\right)$ is:

Let $p\left(x,y\right)$  be any point on the conic. 
$\therefore$ slope of normal $l=-\frac{dx}{dy}$

Equation of normal to conic at $p\left(x,y\right)$ is

 $\left(Y-y\right)=-\frac{dx}{dy}\left(X-x\right)$

$\Rightarrow -y=-\frac{dx}{dy}\left(X-x\right)$     (on x-axis Y = 0)
$\Rightarrow X=x+y\frac{dy}{dx}$  
Given  $X=x+1\Rightarrow y\frac{dy}{dx}=1$

$ydy=dx$
$\Rightarrow \frac{y^2}{2}=x+C.......\left(i\right)$  
 Since the conic passes through origin, x = 0, y = 0 in (i) we get C = 0

$\therefore$ From  $\left(i\right)y^2=2x$
 Which is parabola having length of latus rectum  = $4a=2units$