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).
Best Answer
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 $(a,0)$