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I have the point $(1,-1)$

and the ellipse $$x^2/9 + y^2/5 = 1 $$

How to find the minimum and maximum distance of the point from the ellipse ?

from exploring the ellipse I know that $$a = 3$$ , $$b =\sqrt{5}$$ $$ c = \sqrt{a^2-b^2} =\sqrt{9-5} = \sqrt{4}=2$$

the eccentricity of the ellipse is $$e=c/a = 2/3 $$ the center is $(0,0)$ and the guides are $$x=3,~~x=-3,~~y=\sqrt{5},~~y=-\sqrt{5}$$

the focus points are : $(2,0)$ and $(-2,0)$

how from all of that do I find the requested in the question?

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4 Answers 4

Using $Mathematica$, I have plotted the solution, which corresponds to roots of a quartic equation, which is why I am only going to show you a picture and a numerical approximation of the coordinates, which are $(1.38065, -1.98519)$, and $(-2.84987, 0.698515)$. enter image description here

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I cross-referenced your values with the zeros of my equation above. Everything checks out. Cool cool cool. –  Kaj Hansen Jun 10 '14 at 6:06
@heropup so the question requests me to find the lengths of the red and green lines ? –  Lena Bru Jun 10 '14 at 7:57
The minimum and maximum distances are given by the square roots of the two distinct real roots of the quartic $$4 x^4-136 x^3+1465 x^2-5778 x+4805.$$ These are approximately $1.0561749624077723904$ and $4.2079002058083894257$. –  heropup Jun 10 '14 at 8:12
I actually wanted to know how you got to that equation –  Lena Bru Jun 10 '14 at 10:11

The ellipse can be parametrized as follows: $\alpha(t) = \langle 3\cos(t), \sqrt{5}\sin(t)\rangle$ such that $0 \leq t \leq 2\pi$.

From here, note that finding the points that minimize and maximize the distance will be the same points that minimize/maximize the square of the distance. With this trick, we can eliminate some yucky square roots. Applying the Pythagorean theorem, we can define a function $f$ that represents the square of the distance from $(1, -1)$ to an arbitrary point on the ellipse:

$$f(t) = \Big(1 - 3\cos(t)\Big)^2 + \Big(-1 - \sqrt{5}\sin(t)\Big)^2$$

Computing the derivative of this function, we get:

$$f'(t) = 2\cos(t)\Big(\sqrt{5} - 4\sin(t)\Big) + 6\sin(t)$$

The derivative has $2$ zeros on the interval $[0, 2\pi]$. Those should be the $t$-values that minimize and maximize the distance from your point.

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To find the lengths of the red/green lines in heropup's graphics, simply plug your zeros back into $f(t)$ and take the square root of the values. –  Kaj Hansen Jun 10 '14 at 8:26

One way to do it (the most straightforward way) is to use conditional maxima and minima of a function in two variables using Lagrange multipliers.

i.e do this, take a general point on the ellipse as P(x,y) and given point as A(-1,1) f(x,y) = (square of distance between P and A) Obviously when f is maximum, so is the distance and the same with the minimum. Now write a condition (i.e the equation of the ellipse in implicit form)

Now construct this new function F(x,y,L)=(square of distance)+L(implicit equation of ellipse) Now take 3 partial derivatives with respect to x, y and L, equate them to zero and solve for x and y (L being a parameter now) Then using those, substitute for x and y in your distance function and you're done.

(A slightly better way to do this is to find out the normal to the ellipse passing through (-1,1) but finding the maximum with geometric methods isn't easy.)

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Nice visualization is given by heropup's post.

Find equation to a circle with known center and with red line length as a yet unknown radius $u$.

To find tangent points, solve for either x or y ( only half solution required) between ellipse and red line circle.

Equate discriminant to zero as the two roots coincide. Resulting condition is adequate to find both the tangent points.

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