$P$ satisfies $x^2+(y-4)^2=1$, and $Q$ satisfies $\frac{x^2}{4}+y^2=1$. What is the maximum value of $|PQ|$?
I tried to solve this graphically, but can't sure whether the maximum that I saw is the correct one. What I could see based on the diagram, max of $|PQ|$ is 6, but could it be possible that there is a greater value then 6?
I also thought of differentiation, but the $|PQ|$ function has too many variables. $|PQ|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
Any ideas to solve this by calculation? Thanks
On
I solve this problem in the complex plane as follows: recognizing that the solution must rest in the upper part of the circle and lower half of the ellipse, we can write
$$ z_c=e^{i\theta},\quad \theta\in[0,\pi]\\ \begin{align} z_e &=2\cos \theta+i\sin\theta,\quad \theta\in[\pi,2\pi]\\ &=-2\cos \theta-i\sin\theta,\quad \theta\in[0,\pi]\\ \end{align} $$
Then,
$$ \begin{align} |z_c-z_e| &=|\cos\theta+i(\sin\theta+4)-(-2\cos \theta-i\sin\theta)|\\ &=|3\cos\theta+i(2\sin\theta+4)|\\ &=\sqrt{9\cos^2\theta+4\sin^2\theta+16\sin\theta+16}\\ &=\sqrt{5\cos^2\theta+16\sin\theta+20}\\ \end{align} $$
We can now optimize by setting
$$\frac{d}{d\theta}|z_c-z_e|=\frac{1}{2}\frac{-10\cos\theta\sin\theta+16\cos\theta}{\sqrt{5\cos^2\theta+16\sin\theta+20}}=0$$
The possible solutions are
$$ \sin\theta=\frac{16}{10},\quad \text{impossible}\\ \cos\theta=0,\quad \text{max value at } \theta=\pi/2\\ $$
In other words, from the top of the circle to the bottom of the ellipse, with $|z_c-z_e|=\sqrt{5\cos^2\theta+16\sin\theta+20}=\sqrt{36}=6$.
Suppose $P,Q$ are points, with $P$ on the circle, and $Q$ on the ellipse, such that the distance from $P$ to $Q$ is maximized. Let $w = d(P,Q)$.
From the geometry, it's clear that $P$ must be on the upper half of the circle, and $Q$ must be on the lower half of the ellipse.
As a necessary condition, the line segment $PQ$ must be perpendicular to the circle at the point $P$. It follows that the segment $PQ$ must pass through the center of the circle. Hence, since the radius of the circle is $1$, we must have$\;w=v+1$, where $v$ is the maximum possible distance from the center of the circle,$\;(0,4),\;$to a point $Q$ on the ellipse.
Maximizing the square of the distance will make the job easier$\,-\,$we can take the square root of the result later.
Let $Q=(x,y)$.
By symmetry, we can assume $x \ge 0$, and from the geometry, we must have $-1 \le y \le 0$.
Then $x,y$ are such that $$x^2 + (y-4)^2$$ is maximized, subject to the constraint $$\frac{x^2}{4}+y^2=1$$ Replacing $x^2$ by $4-4y^2$, you want to maximize $$f(y) = (4-4y^2) + (y-4)^2$$ subject to the constraint $-1 \le y \le 0$.
So the problem is reduced to a one-variable maximization of a quadratic polynomial on a closed interval.
Can you finish it?
After you find the maximum value of $f$ on the interval $[-1,0]$, don't forget to take the square root (since we maximized the square of the distance to the center of the circle), and finally, remember that to get the final answer, we have to add $1$ (the radius of the circle).
As it turns out, the final answer, as anticipated, is $6$, with the obvious points $P,Q$, but if you change the term ${\large{\frac{x^2}{4}}}$ to ${\large{\frac{x^2}{a^2}}}$, for some fixed $a > \sqrt{5}$, then using the same solution method as outlined above, the point $Q$ will be on the lower half of the ellipse, but not at a vertex, so the final answer would then be greater than $6$.