Question

The line $x+y=1$ touches the parabola $y^2-y+x=0$ at the point

• A. $(1,1)$
• B. $(\frac{1}{2},\frac{1}{2})$
• C. $(0,1)$
• D. $(1,0)$

Answer 1

The correct option is C $(0,1)$

$y^2-y+x=0{(y-\frac{1}{2})}^2=-(x-\frac{1}{4})$
$\therefore a=\frac{1}{4}$
Let $X=x-\frac{1}{4},Y=y-\frac{1}{2}$
Hence, theparabola will become $Y^2=-X$
Slope of the given line is $m=-1$
$\therefore$ Point of contact
$\equiv (-\frac{a}{m^2},-\frac{2a}{m})\equiv (-\frac{1}{4},\frac{1}{2})$
Now,
$\Rightarrow x-\frac{1}{4}=-\frac{1}{4}\Rightarrow x=0y-\frac{1}{2}=\frac{1}{2}\Rightarrow y=1$
$\therefore$ The point of contact is $(0,1)$