Find shortest distance from point on ellipse to focus of ellipse.

Question

Find the point $(x,y)$ on the ellipse $b^2x^2 + a^2y^2 = a^2b^2$ such that the distance to the focus $(c,0)$ is a minimum.

My book I got this problem out of gave a suggestion saying to express the distance as a function of $x$ and work the problem and then express the distance as a function of $y$ and work the problem.

So, I tried to work it out:

$$x^2 = \frac{a^2b^2 - a^2y^2}{b^2}$$ $$y^2 = \frac{a^2b^2 - b^2x^2}{a^2}$$

Now, I write the distance function:

$$d = \sqrt{(x-c)^2 + y^2}$$

Then, I express the distance function as a function of x:

$$d = \sqrt{(x-c)^2 + \frac{a^2b^2 - b^2x^2}{a^2}}$$

Then, I took the derivative w.r.t $x$:

$$d' = 2x - 2c - \frac{2b^2}{a^2}$$

Now, I did as the book told me, and expressed the distance function as a function of y:

$$d' = -\frac{2a^2}{b^2} + \frac{4a^2c}{bx} + 2y$$

I tried to set both derivatives to zero and attempted to solve, but I couldn't really figure out what to do (my algebra is poor).

Any help would be appreciated. Thanks.

Edit: I need a calculus solution (I'm working out of a calculus book).

Answer 1

The hint.

We need to find a a values of $d>0$, for which the system $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ and $$(x-c)^2+y^2=d^2$$ has solutions.

We obtain: $$\frac{x^2}{a^2}+\frac{d^2-(x-c)^2}{b^2}=1$$ or $$\left(\frac{1}{a^2}-\frac{1}{b^2}\right)x^2+\frac{2cx}{b^2}+\frac{d^2-c^2}{b^2}-1=0$$ or $$-\frac{c^2x^2}{a^2b^2}+\frac{2cx}{b^2}+\frac{d^2-a^2}{b^2}=0$$ or $$c^2x^2-2ca^2x+a^2(a^2-d^2)=0$$ or $$(cx-a^2)^2=a^2d^2,$$ which gives $$x=\frac{a^2+ad}{c},$$ which is impossible or $$x=\frac{a^2-ad}{c}.$$ Now, we need that the equation $$\frac{\left(\frac{a^2-ad}{c}\right)^2}{a^2}+\frac{y^2}{b^2}=1$$ has real roots for which we need $$1-\frac{\left(\frac{a^2-ad}{c}\right)^2}{a^2}\geq0$$ or $$c^2-(a-d)^2\geq0,$$ which gives $$a-c\leq d\leq a+c.$$ Id est, $d\geq a-c$.

The equality occurs for the point $(a,0)$ on the ellipse, which says that $a-c$ it's the answer.

Another way.

Let $F_1(c,0)$, $F_2(-c,0)$, $A(a,0)$ and let $M(x,y)$ be a point on the ellipse.

Thus, $$MF_1+MF_2=2a$$ and by the triangle inequality $$MF_1+F_1F_2\geq MF_2$$ or $$2MF_1+F_1F_2\geq MF_1+MF_2$$ or $$2MF_1+2c\geq2a,$$ which gives $$MF_1\geq a-c.$$ The equality occurs, when $M\equiv A$, which says that $a-c$ is a minimal value.

Answer 2

Since focal distance $ = e \times$distance from directrix ($e=$eccentricity = $\frac{\sqrt{a^2-b^2}}{a}$), we only need to locate points on the ellipse that are closest to a directrix and those would be the corresponding end point of the major axis in this case $(a,0)$

Answer 3

Assuming $a>b$, we have $c=\sqrt{a^2-b^2}$.

Using the parametric equations

$$\begin{cases}x=a\cos\theta,\\y=b\sin\theta\end{cases}$$

we minimize

$$(a\cos\theta-c)^2+(b\sin\theta)^2.$$

The derivative of this expression is

$$-a\sin\theta(a\cos\theta-c)+b\cos\theta(b\sin\theta).$$

There is a root for $$\sin\theta=0$$ (on the major axis) and for

$$(a^2-b^2)\cos\theta=ac,$$ or

$$\cos\theta=\frac a{\sqrt{a^2-b^2}}>1 (!).$$

Hence,

$$d=a-\sqrt{a^2-b^2}.$$