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. \[{x^2} = {a^2} + {y^2}\]
VerifiedHint: 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.
