Question

If $x$ and $y$ co-ordinates of a point $P$ in the $xy$ plane are given by $x=\left(u\cos \alpha \right)t,y=\left(u\sin \alpha \right)t-\frac{1}{2}g{t}^2$ where $t$ is a parameter and $g,u$ and $\alpha$ are given constants. Then the locus of the point $P$ is a parabola whose vertex is

• A. $(\frac{u^2\sin \alpha \cos \alpha }{g},\frac{u^2{\sin }^2\alpha }{2g})$
• B. $(\frac{u^2{\sin }^2\alpha }{2g},\frac{u^2\sin \alpha \cos \alpha }{g})$
• C. $(\frac{u^2\sin \alpha \cos \alpha }{2g},\frac{u^2{\sin }^2\alpha }{g})$
• D. $(\frac{u^2{\sin }^2\alpha }{g},\frac{u^2\sin \alpha \cos \alpha }{2g})$

Answer 1

The correct option is A $(\frac{u^2\sin \alpha \cos \alpha }{g},\frac{u^2{\sin }^2\alpha }{2g})$

Eliminating the parameter $t$, we get
$y=x\tan \alpha -\frac{1}{2}g\left(\frac{x^2}{u^2{\cos }^2\alpha }\right)$

$\Rightarrow x^2-2x\frac{u^2\sin \alpha \cos \alpha }{g}=-\frac{2u^2{\cos }^2\alpha }{g}y$
Adding $\frac{u^4{\sin }^2\alpha {\cos }^2\alpha }{g^2}$ both sides, we get
${(x-\frac{u^2\sin \alpha \cos \alpha }{g})}^2=\frac{-2u^2{\cos }^2\alpha }{g}(y-\frac{u^2{\sin }^2\alpha }{2g})$
It represents a parabola whose vertex is
$(\frac{u^2\sin \alpha \cos \alpha }{g},\frac{u^2{\sin }^2\alpha }{2g})$