Question

The position of a moving point in X-Y plane at time $t$is given by $\left( {u\cos \alpha t,u\sin \alpha t - \dfrac{1}{2}g{t^2}} \right)$ , where $u,\alpha ,g$are constants. What is the locus of point?
A. A circle
B. A parabola
C. An ellipse
D. None of the above

Answer 1

Hint: We will first assume the point to be $\left( {h,k} \right)$.Then we will equate the value of $h$ and $k$ to the given values .By this we will get two equation ,one involving $h$,$u,\alpha $and $t$,second involving $k$,$u,\alpha $and $t$.From both equation we will try to get rid of $t$as it is only variable .To eliminate $t$,we will find the values of $t$in terms of other variables from first equation obtained earlier and substitute this value of $t$in the second equation .After solving the equation hence got ,we can find the locus of point.

Formula Used:
$\tan \alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }}$
$\sec \alpha = \dfrac{1}{{\cos \alpha }}$

Complete step by step solution:
Let the coordinates of point be $\left( {h,k} \right)$.But the coordinates of points are $\left( {u\cos \alpha t,u\sin \alpha t - \dfrac{1}{2}g{t^2}} \right)$.
Hence,
$h = u\cos \alpha t$ (1)
$k = u\sin \alpha t - \dfrac{1}{2}g{t^2}$ (2)
Solving for $t$from equation (1)
$h = u\cos \alpha t$
$ \Rightarrow t = \dfrac{h}{{u\cos \alpha }}$
We know from equation (2)
$k = u\sin \alpha t - \dfrac{1}{2}g{t^2}$
Putting $t = \dfrac{h}{{u\cos \alpha }}$ in the above equation
$k = u\sin \alpha \left( {\dfrac{h}{{u\cos \alpha }}} \right) - \dfrac{1}{2}g{\left( {\dfrac{h}{{u\cos \alpha }}} \right)^2}$
Simplifying the equation
$ \Rightarrow k = h\dfrac{{\sin \alpha }}{{\cos \alpha }} - \dfrac{1}{2}\dfrac{{{h^2}}}{{{u^2}{{\cos }^2}\alpha }}g$
Using the formula$\tan \alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }}$ and $\sec \alpha = \dfrac{1}{{\cos \alpha }}\,\,\,$
$k = h\tan \alpha - \dfrac{1}{2}\dfrac{{{{\sec }^2}\alpha }}{{{u^2}}}g$
Rearranging the terms
$k = \left( {\tan \alpha } \right)h - \left( {\dfrac{{g{{\sec }^2}\alpha }}{{2{u^2}}}} \right){h^2}$
Since $\alpha $,$u$ and$gz$are constant. Let ${k_1} = \tan \alpha $ and ${k_2} = \dfrac{{g{{\sec }^2}\alpha }}{{2{u^2}}}$
Then $k = {k_1}h - {k_2}{h^2}$
To get the locus of point put $h = x$ and $k = y$ in the above equation
$y = {k_1}x - {k_2}{x^2}$
This equation is of the form $y = a{x^2} + bx + c$ where $a = {k_1}$ ,$b = {k_2}$ and $c = 0$,which is the general form of parabola. So, the locus of points is a parabola.

Option ‘B’ is correct

Note: We often make mistake by writing the whole equation in terms $\sin \alpha $ and $\cos \alpha $and eliminating $\alpha $using the formula ${\sin ^2}\alpha + {\cos ^2}\alpha = 1$.This method is wrong because in this case the locus so obtained will contain terms of $t$also but we know that the locus equation must contain any variable other than $x$ and $y$.So here we will eliminate $t$,which is the only variable to get the answer.