A) A circle
B) A parabola
C) An ellipse
D) None of these
Correct Answer: B
Solution :
Let \[h=u\,\cos \,\alpha \,.\,t,\,\,k=u\,\sin \alpha \,.\,t-\frac{1}{2}g{{t}^{2}},\] then \[t=\frac{h}{u\,\cos \alpha }\]. Putting the value of t in \[k=u\,\sin \alpha \,.\,t-\frac{1}{2}g{{t}^{2}},\] we get \[k=h\,\tan \alpha -\frac{1}{2}g\frac{{{h}^{2}}}{{{u}^{2}}{{\cos }^{2}}\alpha }\] \[\therefore \,\,\]Locus of (h, k) is \[y=x\tan \alpha -\frac{1}{2}g\frac{{{x}^{2}}}{{{u}^{2}}\,{{\cos }^{2}}\alpha }\], which is a parabola.
You need to login to perform this action.
You will be redirected in
3 sec
