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Q.
Minimum distance between the curves $y^{2}=4 x$ and $x^{2}+y^{2}-12 x+31=0$ is
ManipalManipal 2013
Solution:
Centre and radius of the given circle is $P(6,0)$ and $\sqrt{5}$, respectively.
Now minimum distance between two curves always occurs along a line which normal to both the curves.
Equation of normal to $y^{2}=4 x$ at $\left(t^{2}, 2 t\right)$ is
$y=-t x+2 t+t^{3}$
If it is normal to circle also, then it must pass though $(6,0)$.
$\therefore 0=t^{3}-4 t$
$\Rightarrow t=0$ or $t=\pm 2$
$\Rightarrow A(4,4)$ and $(4,-4)$
$\Rightarrow P A=P C=\sqrt{20}=2 \sqrt{5}$
$\Rightarrow $ Required minimum distance $2 \sqrt{5}-\sqrt{5}=\sqrt{5}$
