Q89 / 200 · Manipal (MET) 2012 · Mathematics
The curve described parametrically by $x=t^2+t+1, y=t^2-t+1$ represents
- Aa pair of straight lines
- Ban ellipse
- Correct answerCa parabola
- Da hyperbola
Solution
We have, $\quad x=t^2+t+1$ $\qquad \mathrm{...(i)}$
and $\quad y=t^2-t+1$ $\qquad \mathrm{...(ii)}$
Now, $\quad x+y=2\left(1+t^2\right)$ $\qquad \mathrm{...(iii)}$
and $\quad x-y=2 t$ $\qquad \mathrm{...(iv)}$
Now, from Eqs. (iii) and (iv), we get
$\begin{aligned}
& x+y=2\left[1+\left(\frac{x-y}{2}\right)^2\right] \\
\Rightarrow & x+y=2\left[\frac{4+x^2+y^2-2 x y}{4}\right] \\
\Rightarrow & x^2+y^2-2 x y-2 x-2 y+4=0 \qquad \mathrm{...(v)}
\end{aligned}$
$\therefore$ On comparing with
$a x^2+2 h x y+b y^2+2 g x+2 f y+c=0$
We get, $a=1, b=1, c=4, h=-1, g=-1$, $f=-1$
$\because \quad \Delta=a b c+2 f g h-a f^2-b g^2-c h^2$
Now, $\quad \Delta=1 \cdot 1 \cdot 4+2(-1)(-1)(-1)-1 \times(-1)^2$
$\begin{aligned}
& -1 \times(-1)^2-4(-1)^2 \\
& =4-2-1-1-4 \\
& =-4 \text { therefore, } \Delta \neq 0
\end{aligned}$
and $a b-h^2=1 \cdot 1-(1)^2=1-1=0$
So, it is equation of a parabola.
Hence, the correct answer is option: 'C'.