Question

The parametric equation of a parabola is $x = {t^2} + 1$,$y = 2t + 1$. The Cartesian equation of its directrix is
$
  (a){\text{ y = 0}} \\
  (b){\text{ x = - 1}} \\
  (c){\text{ x = 0}} \\
  (d){\text{ x - 1 = 0}} \\
 $

Answer 1

Hint – In this question use one of the parametric equations to find the relation between the variable and t, then substitute the value of this t into another equation, to obtain the equation of parabola. Then use this equation obtained to find the directrix.

Complete step-by-step answer:

Given parametric equation of parabola is
$x = {t^2} + 1$................. (1)
And $y = 2t + 1$...................... (2)
Now from equation (2) the value of t is
$ \Rightarrow t = \dfrac{{y - 1}}{2}$
Substitute this value in equation (1) we have,
$ \Rightarrow x = {\left( {\dfrac{{y - 1}}{2}} \right)^2} + 1$
Now simplify the above equation we have,
$ \Rightarrow x = {\left( {\dfrac{{y - 1}}{2}} \right)^2} + 1$
$ \Rightarrow x = \dfrac{{{{\left( {y - 1} \right)}^2}}}{4} + 1$
$ \Rightarrow {\left( {y - 1} \right)^2} = 4\left( {x - 1} \right)$
So this is the required equation of parabola.
Now compare it with standard form of parabola ${Y^2} = 4aX$
This general equation of parabola is centered at (0, 0).
And equation of directrix is $X = - a$
So on comparing we have,
$Y = y - 1,{\text{ }}X = x - 1{\text{ and }}4a = 4 \Rightarrow a = 1$
So the equation of directrix is
$ \Rightarrow X = - a$
$ \Rightarrow x - 1 = - 1$
$ \Rightarrow x = 0$
So this is the required equation of directrix.
Hence option (C) is correct.

Note – A parabola is a set of all points in a plane which are an equal distance away from a given point and a given line. This point is called the focus and the line is called the directrix. The directrix is perpendicular to the axis of symmetry of a parabola and does not touch the parabola.