Question

Locus of the point of intersection of the lines $x = a{t^2},y = 2at$ is
A. ${x^2} + {y^2} = 2{a^2}$
B. ${y^2} = 4ax$
C. ${x^2} + {y^2} - ax = 0$
D. ${x^2} = {a^2} + {y^2}$

Answer 1

First of all, convert the lines in terms of $t$ and equate them to find the locus of the point of intersection of the given lines. So, use this concept to reach the solution of the given problem.

Complete step-by-step answer:

Given lines are $x = a{t^2}$ and $y = 2at$ can be represented as a point $\left( {a{t^2},2at} \right)$ as shown in the given below:

Which can be rewritten as ${t^2} = \dfrac{x}{a}.....................................\left( 1 \right)$

And $t = \dfrac{y}{{2a}}....................................................\left( 2 \right)$

Substituting equation (2) in (1), we get

$$\Rightarrow {\left( {\dfrac{y}{{2a}}} \right)^2} = \dfrac{x}{a} \\\ \Rightarrow \dfrac{{{y^2}}}{{4{a^2}}} = \dfrac{x}{a} \\\ \Rightarrow {y^2} = \dfrac{{4{a^2}x}}{a} \\\ \therefore {y^2} = 4ax \\\$$

Hence the locus of the point of intersection of the lines $x = a{t^2},y = 2at$ is ${y^2} = 4ax$.

Thus, the correct option is B. ${y^2} = 4ax$.

Note: The formed equation is a standard equation of a parabola which is a conic section having transverse axis (line of symmetry) as x-axis. In this question we have eliminated the variable term $t$ to find the required locus of points of intersection of the given line.