Question

For the curve which is described parametrically by $x=t^2+t$ and $y=t^2-t$, the value of $|\Delta |$ is

Answer 1

Parametric equations are
$x=t^2+t\cdots \left(1\right)y=t^2-t\cdots \left(2\right)$
From $\left(1\right)+\left(2\right),$ we get
$\frac{x+y}{2}=t^2\cdots \left(3\right)$
From $\left(1\right)-\left(2\right),$ we get
$\frac{x-y}{2}=t\cdots \left(4\right)$

Now from $\left(3\right)$ and $\left(4\right),$ we have
$\frac{x+y}{2}={\left(\frac{x-y}{2}\right)}^2\Rightarrow x^2+y^2-2xy-2x-2y=0$
Comparing the above equation with general form of conic equation, we get
$a=b=1,h=-1,g=f=-1,c=0$
$\therefore \Delta =0-2-1-1-0=-4\Rightarrow |\Delta |=4$