Question

The locus of a point whose co-ordinates are curve by $x=t+t^2,y=2t+1$, where $t$ is variable

• A. $y^2=4x+1$
• B. $y^2=4(x+1)$
• C. $y^2=4x$
• D. $Noneofthese$

Answer 1

The correct option is A $y^2=4x+1$

We have,

$x=t+t^2$ and $y=2t+1$ $\Rightarrow \frac{y-1}{2}=t$

Now,

$\begin{array}{c}\Rightarrow x=\frac{y-1}{2}+{\left(\frac{y-1}{2}\right)}^2\\ \Rightarrow x=\frac{2\left(y-1\right)}{4}+\frac{{\left(y-1\right)}^2}{4}\end{array}$

$\begin{array}{c}\Rightarrow 4x=2y-2+y^2+1-2y\\ hence,\\ \Rightarrow 4x+1=y^2\end{array}$

Hence, option $A$ is correct answer.