Minimizing the sum of the distance of two points and a point on an ellipse

Question

Consider coordinates $B=(-16,0)$ and $A=(12,8).$ Let $D=(x,y)$ be a coordinate of the ellipse $9x^2+25y^2=3600$ such that the distance of $AD+DB$ is minimized. For what value of $(x,y)$ is $AD+DB$ minimzed and what will the minimal distance be?

Firstly, I noticed that $(-16,0)$ is one of the foci of the ellipse (the left one).

After doing some graphing and testing values, I noticed that the distance of $AD+DB$ was minimized when $\angle ADB$ was a right angle. (Does this hold true for all ellipses? If so, why does it?). It seems like I have some information, but I'm not sure how to apply it to solve the problem. If $AD+DB$ is minimzed in this configuration, I would still need the coordinate of $D$ to find $AD$ and also $\sqrt{BA^2 + AD^2}=DB.$ However, I'm not fully sure how to do this. Thanks in advance.

Edit: I find that $AD+BD$ is approximately $29.18+1.9,$ which is close to $31.$ However, would there be a mathematical way to do this and get a more precise answer?

Answer 1

There is a simple geometrical solution. Sum of distances is minimum when the ellipse of foci $A$, $B$ passing through $D$ is tangent to the given ellipse. Hence both ellipses must have the same normal at $D$, i.e. angles $BDA$ and $BDC$ must have the same bisector and that implies they must coincide. Hence point $D$ gives a minimum sum $AD+DB$ if points $CAD$ are aligned.

Answer 2

First of all, it's important to identify $B(-16,0)$ and $C(16,0)$ with $f=16$ (formula $a^2=b^2+f^2$) as the foci of the ellipse.

Have a look at this figure:

Among the ellipses having the common foci $A$ and $B$, consider the unique one (in red) which is internaly tangent to the original ellipse, therefore sharing a common tangent (and a common normal) with the latter.

Now apply the optical property of the ellipse (all rays emitted by a focus are reflected so as to converge to the other focus) to both ellipses, don't you think that point $M$ and point $C$ must coincide ?

Otherwise said, $D$ must be aligned with $A$ and $C$.

Now it remains to express this analytically: you will check that straight line $AD$ has equation:

$$y=-2x+32$$

Plugging this expression into the equation

$$9x^2+25y^2=3600$$

of the ellipse gives a quadratic equation yielding the 2 abscissas of intersection of line $AC$ with the ellipse, i.e., the abscissa of $D$. Take any of them (in fact both are valid !) and replace $x$ by its value in (1) to obtain the corresponding ordinate $y$.

Once you have the coordinates $(x,y)$ of point $D$, you have to compute

$$DA+DB=\sqrt{(x-12)^2+(y-8)^2}+\sqrt{x^2+(y-10)^2}$$

in a tedious way; I don't see any shortcut.

Answer 3

This approach might be a lengthy one, but it will give you the accurate answer. There are other simple solutions to this, but this is how you would do it by single-variable calculus:

---

Assumption: $D$ lies in the first quadrant (assumption was motivated from the graph drawn)

The most straightforward approach to this question will be with the help of differential calculus.

We start off by slightly changing the equation of the ellipse like so:

$$9x^2+25y^2=3600$$ $$\frac{x^2}{400}+\frac{y^2}{144}=1$$

Recall that the parameterization for a standard ellipse: $\frac{x^2}{a^2}+\frac{y^2}{b^2}$ is: $(a\cos(t),b\sin(t))$

So for the given ellipse, we can generalize any point on it as: $(20\cos t,12\sin t). $

Given we need to minimize the distance $AD+DB$, We will ultimately have to minimize the function: $$\boxed{f(t)=\sqrt{(20\cos t-12)^2+(12\sin t-8)^2}+\sqrt{(20\cos t+16)^2+(12\sin t)^2}}$$

---

Computing $f'(t)$, we get: $$\dfrac{24\cos\left(t\right)\left(12\sin\left(t\right)-8\right)-40\left(20\cos\left(t\right)-12\right)\sin\left(t\right)}{2\sqrt{\left(12\sin\left(t\right)-8\right)^2+\left(20\cos\left(t\right)-12\right)^2}}+\dfrac{288\cos\left(t\right)\sin\left(t\right)-40\left(20\cos\left(t\right)+16\right)\sin\left(t\right)}{2\sqrt{144\sin^2\left(t\right)+\left(20\cos\left(t\right)+16\right)^2}}$$

On finding the zeroes of the derivative above, we have: $$t=2\left(n\pi+\arctan\left(\frac 16(1+\sqrt{5})\right)\right)$$ Now we check on a few values of $n$ to get the minimum distance.

Answer 4

We can simply apply triangle inequality. You know that sum of distance to any point on ellipse from both focii is $2a$ which is length of major axis.

$B$ and $C$ are focii. Using triangle inequality $DA + AC \geq DC$

$BD + DA \geq BD + (DC - AC)$
i.e $BD + DA \geq 2a - AC$

Please note $AC$ is fixed given points $A$ and $C$ are fixed and so is $2a$ which is $40$. So $BD + DA$ is minimum at equality i.e. when $DA + AC = DC$ which occurs when point $D$ is such that $A$ is on line segment $DC$.

$2a = 40, AC = \sqrt{80}$. So minimum value of $BD + DA = 40 - \sqrt{80}$