Question

A particle moves in $xy-$plane according to the equation $x=\alpha t$ and $y=\alpha t(1-\beta t)$ where $\alpha$ and $\beta$ are positive constants and $t$ is time.

Choose the correct option in regard to the trajectory of the particle.

• A. Equation of the trajectory of the particle is $y=x(1+\frac{\beta }{\alpha }x)$
• B. Equation of the trajectory of the particle is $y=x(1+\frac{\alpha }{\beta }x)$
• C. Trajectory of the particle is symmetric about $x=\frac{2\alpha }{\beta }$
• D. Trajectory of the particle is symmetric about $x=\frac{\alpha }{2\beta }$

Answer 1

The correct option is D Trajectory of the particle is symmetric about $x=\frac{\alpha }{2\beta }$

Given:

$x=\alpha t\Rightarrow t=\frac{x}{\alpha }...\left(i\right)$, and,

$y=\alpha t(1-\beta t)$

$\Rightarrow y=\alpha \times \frac{x}{\alpha }(1-\beta \times \frac{x}{\alpha })\left[\text{From}\right(i\left)\right]$

$\Rightarrow y=x(1-\frac{\beta }{\alpha }x)$
$[$Equation of the trajectory of the particle$]$

Now, the above expression can be written as,

$y=x-\frac{\beta }{\alpha }{x}^2$

Comparing this with standard form of quadratic equation,

$y=a{x}^2+bx+c$

We get,
$a=-\frac{\beta }{\alpha },b=1,$
$c=0$

We know that the axis of symmetry of a quadratic equation is given as $x=\frac{-b}{2a}$

So, the line of symmetry for the given function will be,

$x=\frac{-b}{2a}=\frac{-1}{2(-\frac{\beta }{\alpha })}=\frac{\alpha }{2\beta }$

Hence, trajectory of particle is symmetric about $x=\frac{\alpha }{2\beta }$