Question

In the graph of the parametric equations $\{\begin{array}{c}x=t^2+1\\ y=t^2-1\end{array}$

• A. $x\ge 0$
• B. $x\ge -\frac{1}{4}$
• C. x is any real number
• D. $x\ge -1$
• E. $x\le 1$

Answer 1

Answer: (B)

$x\ge -\frac{1}{4}$

Given parametric equations

$x=t^2+t$

$y=t^2-t$

For graph formed using above parametric equations 't' is always Real.

$\therefore$Discrimination of $t^2+t-x=0$ must be $\ge 0$

$\therefore {\left(1\right)}^2-4\left(1\right)(-x)\ge 0$

$1+4x\ge 0$

$\therefore x\ge \frac{-1}{4}$