Question

If the line $x+y-1=c$ touches the parabola $x^2+y-x=0,$ then the value of $c$ is

Answer 1

$x^2+y-x=0\Rightarrow {(x-\frac{1}{2})}^2=-(y-\frac{1}{4})$
Now,
$Y=y-\frac{1}{4},X=x-\frac{1}{2}\therefore X^2=-Y\therefore a=\frac{1}{4}$
Now,
$x+y=c+1\Rightarrow X+Y=c+\frac{1}{4}$
Now using condition of tangency
$c+\frac{1}{4}=a{m}^2\Rightarrow c+\frac{1}{4}=\frac{1}{4}\Rightarrow c=0$