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?
On
For sure, as Muralidharan answered, the parametrization of the curve makes things simple.
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 $$\frac{d(D^2)}{dx}=2 \left(x-\frac{1}{3}\right)+\frac{4 \left(x^{2/3}-2\right)}{3 \sqrt[3]{x}}=\frac{2}{3} \left(3 x+2 \sqrt[3]{x}-\frac{4}{\sqrt[3]{x}}-1\right)=0$$ Let $x=t^3$ which makes$$3 t^3+2 t-\frac{4}{t}-1=0$$ Muralidharan wrote and solved.
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 $\frac{\sqrt{5}}{3}$