The coordinates of a point on the parabola ${\mathrm{y}}^2=8\mathrm{x}$ whose distance from the circle ${\mathrm{x}}^2+{(\mathrm{y}+6)}^2=1$ is minimum is

Any point on the parabola ${\mathrm{y}}^2=8\mathrm{x}(4\mathrm{a}=8$ or $\mathrm{a}=2)$ is $\left({\mathrm{at}}^2, 2\mathrm{at}\right)$ or $(2{\mathrm{t}}^2, 4\mathrm{t})$ its minimum distance from the circle means its distance from the center, (0,-6) of the circle.

Let D be the distance. Then

$$\begin{gathered} \mathrm{z}={\mathrm{D}}^2={\left(2{\mathrm{r}}^2\right)}^2+{(4\mathrm{t}+6)}^2=4({\mathrm{t}}^4+4{\mathrm{r}}^2+12\mathrm{t}+9) \\ \therefore \frac{\mathrm{dz}}{\mathrm{dt}}=4(4{\mathrm{t}}^3+8\mathrm{t}+12)=0 \end{gathered}$$

or $16({\mathrm{r}}^3+2\mathrm{t}+3)=0$

or $16(\mathrm{t}+1)({\mathrm{t}}^2-\mathrm{t}+3)=0$

or $\mathrm{t}=-1$

$\frac{{\mathrm{d}}^2\mathrm{z}}{{\mathrm{dt}}^2}=16(3{\mathrm{t}}^2+2)=+\mathrm{ve}.$ Hence, minimum.

Therefore, point is (2,-4)