MCQ
The curve describes parametrically by $x = t^2 - 2t + 2, y = t^2 + 2t + 2$ represents
- Astraight line
- Bpair of straight lines
- Ccircle
- ✓parabola
and $y=t^{2}-t+1$
Now, $x+y=2\left(1+t^{2}\right)$
and $x-y=2 t$
Now, from Eqs. (iii) and (iv), we get
$x+y=2\left[1+\left(\frac{x-y}{2}\right)^{2}\right]$
$\Rightarrow x+y=2\left[\frac{4+x^{2}+y^{2}-2 x y}{4}\right]$
$\Rightarrow x^{2}+y^{2}-2 x y-2 x-2 y+4=0 \quad \ldots(v)$
On comparing with, we get $a x^{2}+2 h x y+b y^{2}+2 g x+2 f y+c=0$
We get, $a=1, b=1, c=4, h=-1, g=-1, f=-1$
$\Delta=a b c+2 f g h-a f^{2}-b g^{2}-c h^{2}$
Now, $\Delta=1 \cdot 1 \cdot 4+2(-1)(-1)(-1)-1 \times(-1)^{2}-1 \times(-1)^{2}-4(-1)^{2}$
$=4-2-1-1-4$
$=-4,$ therefore, $\Delta \neq 0$
and $a b-h^{2}=1 \cdot 1-(1)^{2}=1-1=0$
So, it is equation of a parabola.
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