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The point of contact of the tangent $18 x-6 y+1=0$ to the parabola $y^2=2 x$ is
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The correct answer is:
$\left(\frac{1}{18}, \frac{1}{3}\right)$
Let point of contact be $(h, k)$, then tangent at this point is $k y=x+h$.
$x-k y+h=0 \equiv 18 x-6 y+1=0$
or $\frac{1}{18}=\frac{k}{6}=\frac{h}{1} \quad$ or $\quad k=\frac{1}{3}, \quad h=\frac{1}{18}$
$x-k y+h=0 \equiv 18 x-6 y+1=0$
or $\frac{1}{18}=\frac{k}{6}=\frac{h}{1} \quad$ or $\quad k=\frac{1}{3}, \quad h=\frac{1}{18}$
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