Question

The shortest distance between the curves $y^2=x^3$ and $9x^2+9y^2-30y+16=0$ is

• A. $\frac{\sqrt{13}}{3}$
• B. $\frac{2}{3}$
• C. $\frac{5}{3}$
• D. $\frac{\sqrt{13}}{3}-1$

Answer 1

The correct option is D $\frac{\sqrt{13}}{3}-1$

Given curves are $y^2=x^3$ and $9x^2+9y^2-30y+16=0$
$9x^2+9{(y-\frac{5}{3})}^2=9\Rightarrow x^2+{(y-\frac{5}{3})}^2=1$
Let any point on $y^2=x^3$ be $D(t^2,t^3)$


Now, distance between $C$ and $D$ is
$CD=\sqrt{{(t^2-0)}^2+{(t^3-\frac{5}{3})}^2}\Rightarrow L=C{D}^2=t^4+t^6-\frac{10\times t^3}{3}+\frac{25}{9}$
Differentiating w.r.t. $t$, we get
$\frac{dL}{dt}=6t^5+4t^3-10t^2$
For maxima/minima,
$\frac{dL}{dt}=0\Rightarrow 6t^5+4t^3-10t^2=0\Rightarrow t^2(3t^3+2t-5)=0\Rightarrow t^2(t-1)(3t^2+3t+5)=0\Rightarrow t=0,1(\because 3t^2+3t+5>0\forall t\in R)$
When $t=0,$ $L=\frac{25}{9}$
When $t=1,$ $L=\frac{13}{9}$

Hence, the shortest distance $=\sqrt{\frac{13}{9}}-r=\frac{\sqrt{13}}{3}-1$