Question

The parametric equation of a parabola is $x={{t}^{2}}+1,y=2t+1$. The cartesian equation of its directrix is
A. $x=0$
B. $x+1=0$
C. $y=0$
D. None of these

Answer 1

Hint: We will substitute the value of t as $\left( \dfrac{y-1}{2} \right)$ in the first equation and simplify the equation accordingly and try to get an equation like any standard equation of parabola, which is, ${{y}^{2}}=4ax$. Then we will make use of the fact that the equation of the directrix of the parabola ${{y}^{2}}=4ax$ is given by $x=-a$.

Complete step by step solution:
It is given in the question that the parametric equation of a parabola is $x={{t}^{2}}+1,y=2t+1$ and we have been asked to find the cartesian equation of its directrix. We have been given two equations,
$\begin{align}
  & x={{t}^{2}}+1\ldots \ldots \ldots \left( i \right) \\
 & y=2t+1\ldots \ldots \ldots \left( ii \right) \\
\end{align}$
We can get the value of t form equation (ii) as follows,
$t=\dfrac{\left( y-1 \right)}{2}$
Now, we will put this obtained value of $t=\dfrac{\left( y-1 \right)}{2}$ in equation (i). So, we get,
$\begin{align}
  & x={{\left\{ \dfrac{\left( y-1 \right)}{2} \right\}}^{2}}+1 \\
 & \Rightarrow x=\dfrac{{{\left( y-1 \right)}^{2}}}{4}+1 \\
\end{align}$
On transposing 1 from the right hand side or the RHS to the left hand side or the LHS, we get,
$\left( x-1 \right)=\dfrac{{{\left( y-1 \right)}^{2}}}{4}$
We will now multiply both the sides with 4. So, we get,
$4\left( x-1 \right)={{\left( y-1 \right)}^{2}}$
Now, let us assume $y=\left( y-1 \right)$ and $x=\left( x-1 \right)$. So, we will get,
${{y}^{2}}=4x$
Now, if we compare the general equation of the parabola, ${{y}^{2}}=4ax$ with our obtained equation, ${{y}^{2}}=4x$, then we say that the value of $a=1$. The equation of the directrix is given by $x=-a$, therefore we get the directrix of the given equation as,
$\begin{align}
  & \left( x-1 \right)=-1 \\
 & \Rightarrow x=-1+1 \\
 & \Rightarrow x=0 \\
\end{align}$
Hence, we get the value of $x=0$.
Therefore, the correct answer is option A.

Note: Many times the students take the value of $t=\dfrac{\left( y+1 \right)}{2}$ and as a result they may get the final equation as $4\left( x-1 \right)={{\left( y+1 \right)}^{2}}$, which is incorrect. The next mistake is that they might consider the equation of the directrix as $x=a$ and end up with the value of x as 2 and choose option D as the correct answer. So, the students should remember the basic equation of parabola, ${{y}^{2}}=4ax$.