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If the straight line $x+y=1$ touches the parabola $y^2-y+x=0$, then the co-ordinates of the point of contact are
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The correct answer is:
$(0,1)$
$m$ of tangent $=-1$
Also from equation of parabola, we get gradjent at ${ }^{(h, k)}$ as the slope of parabola
$=\frac{d y}{d x}=\frac{-1}{2 y-1}=\frac{-1}{2 k-1}$
Since line and parabola touch at $(h, k)$
$\Rightarrow \frac{-1}{2 k-1}=-1 \Rightarrow-2 k+1=-1 \Rightarrow k=1$
Putting this value in $x+y=1$, we have $h=0$, so the point of contact is $(0,1)$.
Also from equation of parabola, we get gradjent at ${ }^{(h, k)}$ as the slope of parabola
$=\frac{d y}{d x}=\frac{-1}{2 y-1}=\frac{-1}{2 k-1}$
Since line and parabola touch at $(h, k)$
$\Rightarrow \frac{-1}{2 k-1}=-1 \Rightarrow-2 k+1=-1 \Rightarrow k=1$
Putting this value in $x+y=1$, we have $h=0$, so the point of contact is $(0,1)$.
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