Question

The position of a moving point in $x-y$ plane at time $t$ is given by ($u\cos \alpha .t,u\sin \alpha .t-\frac{1}{2}g{t}^2)$), where $u,g,\alpha$ are constants. If the locus of the moving point is of the form $y=x\tan \alpha -A{x}^2{\sec }^2\alpha$, then find $A$.

Answer 1

We have $x=u\cos \alpha .t$

$\Rightarrow t=\frac{x}{u\cos \alpha }$ .........$\left(1\right)$

$y=ut\sin \alpha -\frac{1}{2}g{t}^2$

$\Rightarrow y=u\sin \alpha .\left(\frac{x}{u\cos \alpha }\right)-\frac{1}{2}g{\left(\frac{x}{u\cos \alpha }\right)}^2$

$\Rightarrow y=\frac{x\sin \alpha }{\cos \alpha }-\frac{1}{2}g{x}^2\left(\frac{1}{u^2{\cos }^2\alpha }\right)$

$\Rightarrow y=x\tan \alpha -\frac{1}{2u^2}g{x}^2{\sec }^2\alpha$

As above equation is linear in $y$ and quadratic in $x$, hence it is an equation of parabola.

Comparing with $y=x\tan \alpha -A{x}^2{\sec }^2\alpha$ we get $A=\frac{1}{2u^2}g$