Question

Vertex of the parabola whose parametric equation is $x=t^2-t+1,y=t^2+t+1;t\in R,$ is

• A. $(1,1)$
• B. $(2,2)$
• C. $(\frac{1}{2},\frac{1}{2})$
• D. $(3,3)$

Answer 1

The correct option is A $(1,1)$

$x=t^2-t+1,y=t^2+t+1$
$x+y=2(t^2+1)\text{and}y-x=2t$
$\Rightarrow \frac{x+y}{2}=1+{\left(\frac{y-x}{2}\right)}^2$
$\Rightarrow (y-x{)}^2=2(x+y)-4$
$\Rightarrow (y-x{)}^2=2(x+y-2)$
Vertex will be the point where the lines
$y-x=0\text{and}x+y-2=0$ meets,
$x+x-2=0\Rightarrow x=1\Rightarrow y=1$
Hence the vertex is $(1,1)$