Question

Find the shortest distance between the curve $y^2=x^3$ and $9x^2+9y^2-30y+16=0$.?

Answer 1

Note that $9x^2+9y^2-30y+16=0$ represents the circle ${(x-0)}^2+{(y-\frac{5}{3})}^2=1^2$ centered at $(0,\frac{5}{3})$ with radius $1$.

The minimum distance between the circle and the second curve $y^2=x^3$.is the minimum distance of the second curve from the center of the circle minus one

This minimum distance is unaltered if the origin is shifted to the point$(0,-\frac{5}{3})$ (by parallel shifting).

By this shifting the equation of the circle reduces to $x^2+y^2=1$ (the unit circle) whose center is the origin $(0,0)$ and the equation of the second curve becomes ${(y+\frac{5}{3})}^2=x^3$.

Hence a typical point of these cond equation can be written in terms of a parameter t as $(x,y)(t^2,t^3-\frac{5}{3})$

Therefore,the square of the distance of the point $(t^2,t^3-\frac{5}{3})$ to the origin is $s\left(t\right)=t^4+{(t^3-\frac{5}{3})}^2=t^4+t^6-\frac{10}{3}{t}^3+\frac{25}{9}$

The minimum value of s corresponds to there solution of the equation $\frac{ds}{dt}=0$

$t^2.(3t^3+2t-5)=0$ which gives two real solutions $t=0,t=1$.

The corresponding points in these cond curve are$(0,-\frac{5}{3})$and$(1,-\frac{2}{3})$.

The corresponding values of s are $\frac{25}{9}$and$\frac{13}{9}$

The minimum distance from the origin to these cond curve is $\frac{\sqrt{13}}{3}$.

Hence the minimum distance between the given circle and these cond curve is $\frac{\sqrt{13}}{3}-1=0.20185$ approximately.