MCQ
The curve described parametrically by $x = {t^2} + t + 1$, $y = {t^2} - t + 1$ represents
- AA pair of straight lines
- BAn ellipse
- ✓A parabola
- DA hyperbola
Hence, $x + y = 2({t^2} + 1)$ and $x - y = 2t$
$\frac{{x + y}}{2} = {t^2} + 1 = {\left( {\frac{{x - y}}{2}} \right)^2} + 1$
$ \Rightarrow $ $2(x + y) = {(x - y)^2} + 4$
$ \Rightarrow $${x^2} - 2xy + {y^2} - 2x - 2y + 4 = 0$
On comparing with $a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$
$a = 1,\,h = - 1,\,b = 1,\,g = - 1,\,f = - 1,\,c = 4$
$\therefore $ $abc + 2fgh - a{f^2} - b{g^2} - c{h^2}$
$ = 1.1.4 + 2( - 1)\,( - 1)\,( - 1)\, - 1.1 - 1.1 - 1.4 = 4 - 2 - 2 - 4 \ne 0$
and ${h^2} = ab$. Hence curve is a parabola.
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