Minimize the ratio involving the ellipse

Question

Let $P$ be any point on the curve $\dfrac{x^2}{4}+\dfrac{y^2}{3}=1$, and $A,B$ be two fixed points $\left(\dfrac{1}{2},0\right)$ and $(1,1)$ respectively. Find the minimum value of $\dfrac{|PA|^2}{|PB|}$.

Assume $x=2\cos\theta,y=\sqrt{3}\sin\theta$, then $$\dfrac{|PA|^2}{|PB|}=\frac{\left(2\cos\theta-\frac{1}{2}\right)^2+(\sqrt{3}\sin\theta-0)^2}{\sqrt{\left(2\cos\theta-1\right)^2+(\sqrt{3}\sin\theta-1)^2}},$$ but which is not so easy to tackle.

WolframeAlpha gives the result $1$.

Answer 1

Doing what @Paul Sinclair already commented and using multiple angle formulae $$\dfrac{|PA|^4}{|PB|^2}=\frac{-256 \cos (\theta )+92 \cos (2 \theta )-16 \cos (3 \theta )+2 \cos (4 \theta )+259}{8 \left(-4 \sqrt{3} \sin (\theta )-8 \cos (\theta )+\cos (2 \theta )+11\right)}$$ Computing the derivative, its numerator is "just" $$(15-8 \cos (\theta )+2 \cos (2 \theta )) \times$$ $$\left(24 \sqrt{3}-40 \sin (\theta )+45 \sin (2 \theta )-12 \sin (3 \theta )+\sin (4 \theta )-40 \sqrt{3} \cos (\theta )-8 \sqrt{3} \cos (2 \theta )+6 \sqrt{3} \cos (3 \theta )\right)$$ The first factor does not cancel and with all these $\sqrt{3} $ in the remaining term, if the solution is simple, it is a multiple of $\frac \pi 3$ and, just trying, one zero of the derivative is $\theta=\frac {5\pi} 3$ and then the result.

Answer 2

Denote $c = \cos \theta$ and $s = \sin\theta$.

We have $$|\mathrm{PA}|^2 = (2c - 1/2)^2 + (s\sqrt 3 - 0)^2= c^2 - 2c + \frac{13}{4}$$ and $$|\mathrm{PB}|^2 = (2c - 1)^2 + (s\sqrt 3 - 1)^2 = c^2 - 4c + 5 - 2\sqrt 3\, s.$$

We have \begin{align*} |\mathrm{PA}|^4 - |\mathrm{PB}|^2 &= (c^2 - 2c + 13/4)^2 - (c^2 - 4c + 5 - 2\sqrt 3\, s)\\ &= (c^2 - 2c + 13/4)^2 - c^2 + 4c - 5 + 2\sqrt 3\, s\\ &\ge 0 \end{align*} where we have used $(c^2 - 2c + 13/4)^2 - c^2 + 4c - 5 > 0$ and \begin{align*} &[(c^2 - 2c + 13/4)^2 - c^2 + 4c - 5]^2 - (2\sqrt 3\, s)^2\\ =\,& [(c^2 - 2c + 13/4)^2 - c^2 + 4c - 5]^2 - 12(1-c^2)\\ =\,&\frac{1}{256}(64c^6 - 448c^5 + 1776c^4 - 4128c^3 + 6524c^2 - 6236c + 4849)(2c-1)^2\\ \ge\,& 0. \end{align*}

On the other hand, when $\theta = \frac{5\pi}{3}$, we have $|\mathrm{PA}|^4 - |\mathrm{PB}|^2 = 0$ and $\frac{|\mathrm{PA}|^2}{|\mathrm{PB}|} = 1$.

Thus, the minimum of $\frac{|\mathrm{PA}|^2}{|\mathrm{PB}|}$ is $1$.