Question

The co-ordinate of a moving point \[P\left( a{{t}^{2}},2at \right)\] where t is a parameter, then find the locus of the point P.

Answer 1

Hint:Take the co-ordinate of point P as (x,y). Thus substitute the value of t in the expression of x. Thus find the locus of point P which will be the standard form of parabola.

Complete step-by-step answer:
We have been given the coordinate of a moving point as \[P\left( a{{t}^{2}},2at \right)\].
Now let us consider the co-ordinate of the point P as (x,y).
Now let us consider that (x,y) = \[\left( a{{t}^{2}},2at \right)\].
Thus we can write from the above that,
\[\begin{align}
  & x=a{{t}^{2}}.........(1) \\
 & y=2at........(2) \\
\end{align}\]
Let’s modify equation (2).
\[\begin{align}
  & y=2at \\
 & \therefore t={}^{y}/{}_{2a} \\
\end{align}\]
Now let us substitute the value of tin equation (1).
\[\begin{align}
  & x=a{{t}^{2}} \\
 & x=a{{\left( \dfrac{y}{2a} \right)}^{2}} \\
 & x=a\dfrac{{{y}^{2}}}{4{{a}^{2}}} \\
\end{align}\]
Cancel out a from the numerator and the denominator.
\[x=a\dfrac{{{y}^{2}}}{4{{a}^{2}}}\]
Now cross multiply the above expression, we get,
\[\begin{align}
  & 4ax={{y}^{2}} \\
 & \Rightarrow {{y}^{2}}=4ax \\
\end{align}\]
This is the locus on the point P.
The equation of locus of point P \[\Rightarrow {{y}^{2}}=4ax\].

Note:We know that \[{{y}^{2}}=4ax\] is the standard form of a parabola, which has a vertex at point (0,0) and the co-ordinates of the focus are (a,0). Thus the equation of the directrix is x =a and the equation of the axis is y = 0.