STD 11 - 10.2 parabola,ellipse,hyperbola — Maths STD 11 Science — Question

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.