Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

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

share|improve this answer
    
I cross-referenced your values with the zeros of my equation above. Everything checks out. Cool cool cool. –  Kaj Hansen Jun 10 at 6:06
    
@heropup so the question requests me to find the lengths of the red and green lines ? –  Lena Bru Jun 10 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 at 8:12
    
I actually wanted to know how you got to that equation –  Lena Bru Jun 10 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.

share|improve this answer
    
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 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.)

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.