What's meant by radius vector of a curve?

Question

Find the length of the minimum radius vector of the curve $\frac{a^2}{x^2}+\frac{b^2}{y^2}=1$.

Let $r$ be the radius vector. So, $x^2+y^2=r^2$. Taking out the value of $x^2$ from the given curve and putting it here, we get value of $r$ in terms of $y$. Taking derivative and equating it to zero, we get the value of $y$ and $x$. Thus, $r$ comes out to be $a+b$.

But what's meant by radius vector of a curve? We are calculating its minima. So, can it have other values as well?

Answer 1

I'm not sure @B.Goddard's comment is right. Here's what I took it to mean:

Just as a point on a curve has a tangent to the curve that is a line agreeing with the curve on $y$ and $y^\prime$ there, you can also choose a circle meeting the curve at a given point so that the circle and curve agree on $y,\,y^\prime,\,y^{\prime\prime}$. The term radius of curvature determines to this circle's radius, so that's the quantity you're asked to minimize here. Presumably, the radius vector is from said circle's centre to the point it touches the curve.

If $\tfrac{a^2}{x^2}+\tfrac{b^2}{y^2}=1$ then$$\begin{align}y^\prime&=-\tfrac{a^2y^3}{b^2x^3}\\\implies y^{\prime\prime}&=-\tfrac{3a^2y^2}{b^2x^3}y^\prime+\tfrac{3a^2y^3}{b^2x^4}\\&=\tfrac{3a^2y^3(a^2y^2+b^2x^2)}{b^4x^6}\\\implies r&=\tfrac{(1+y^{\prime2})^{3/2}}{y^{\prime\prime}}\\&=\tfrac{(a^4y^6+b^4x^6)^{3/2}}{3a^2b^2x^3y^3(a^2y^2+b^2x^2)}.\end{align}$$The parameterization $x=a\sec t,\,y=b\csc t$ gives$$r=\tfrac{(a^2\sec^6t+b^2\csc^6t)^{3/2}}{3a^3b^3\sec t\csc t},$$which shouldn't be hard to minimize.

Answer 2

Let $x=r \cos \theta, y=r \sin \theta$, then $$\frac{r^2\cos^2 \theta}{a^2}+\frac{r^2 \sin^2 \theta}{b^2}=1$$ $$\implies r=\sqrt{\frac{1}{a^{-2} \cos^2 \theta + b^{-2} \sin^2 \theta}}>0,$$ is the radius vector for this ellipse.