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,

& x=a{{t}^{2}}.........(1) \\\ & y=2at........(2) \\\ \end{aligned}$$ Let’s modify equation (2). $$\begin{aligned} & y=2at \\\ & \therefore t={}^{y}/{}_{2a} \\\ \end{aligned}$$ Now let us substitute the value of tin equation (1). $$\begin{aligned} & x=a{{t}^{2}} \\\ & x=a{{\left( \dfrac{y}{2a} \right)}^{2}} \\\ & x=a\dfrac{{{y}^{2}}}{4{{a}^{2}}} \\\ \end{aligned}$$ Cancel out a from the numerator and the denominator. $$x=a\dfrac{{{y}^{2}}}{4{{a}^{2}}}$$ Now cross multiply the above expression, we get, $$\begin{aligned} & 4ax={{y}^{2}} \\\ & \Rightarrow {{y}^{2}}=4ax \\\ \end{aligned}$$ This is the locus on the point P. The equation of locus of point P $$\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.