# Minium distance of curve from given point

Let $d$ be the minimum distance of $y^3=x^2$ from the point $(1/3,2)$ measured in first quadrant, then find the value of $9 d^2$

I was first trying to solve it taking point $(x,x^{2/3})$ on curve using distance formula but it resulted in tedious calculation. Can we say that minimum distance will be obtained along the normal to the curve which passes through $(1/3,2)$? If so then how are supposed to use that to get required distance?

• you can calculate distance and try to minimize it – Nebo Alex Sep 15 '16 at 4:40

The curve can be parametrised by $x=t^3, y=t^2$. The normal at point $t$ is $$y-t^2 = -\frac{3t}{2}(x-t^3)$$ and this passes through $(1/3,2)$ when $t$ satisfies $$3t^4+2t^2-t-4 = 0$$ and this can be factorized as $$(t-1)(3t^3+3t^2+5t+4) = 0$$ and the only root that lies in the first quadrant is $t=1$. Thus the point is $(1,1)$ and the minimum distance is $\frac{\sqrt{5}}{3}$
However, the very basic method is not so tedious if you take into account the fact that minimizing the distance is the same as minimizing its square. So, let us write $$D^2=\left(x-\frac{1}{3}\right)^2+(y-2)^2=\left(x-\frac{1}{3}\right)^2+\left(x^{2/3}-2\right)^2$$ Now, compute the derivative of the above with respect to $x$ to get $$\frac{d(D^2)}{dx}=2 \left(x-\frac{1}{3}\right)+\frac{4 \left(x^{2/3}-2\right)}{3 \sqrt{x}}=\frac{2}{3} \left(3 x+2 \sqrt{x}-\frac{4}{\sqrt{x}}-1\right)=0$$ Let $x=t^3$ which makes$$3 t^3+2 t-\frac{4}{t}-1=0$$ Muralidharan wrote and solved.