I have tried for awhile to get this problem without success. "An ellipse is drawn with major and minor axes of lengths 10 and 8 respectively. Using one focus as center, a circle is drawn that is tangent to the ellipse, with no part of the circle being outsdie the ellipse. Compute the radius of the circle"

One method of using the definition of an ellipse (the focii definition) came up with one focus= (root 18, 0), While common sense tells me when x=0, the focus clearly must =3 since the sum of the distances from the focii is ten and you are splitting this, so you have 3,4,5 triangle.

I also noted that you have an inscriptable quadrilateral formed by the origin, a minor axes point, the center of the circle (the focus), and the point of tangency. Would that be right? I am wondering if you could use any theorems then such as Ptolemys which I can't remember...

Thanks for your help!

Well, the foci are certainly 3 units either side of the centre on the long axis; I think you must have got an extra factor of 2 from somewhere in your calculation. Your common sense is quite correct.

As for the circle, it's really not as difficult as it might seem: the circle you want is tangent to the ellipse at the endpoint of the major axis.

Can you prove that this circle lies wholly within the ellipse? (You'll need the triangle inequality.)

David

(An interesting question to ask next is the size of the largest circle tangent to the ellipse at the end of the major axis which is wholly contained within the ellipse. I think the largest one in this case has radius 16/5, and generally b² /2a where the ellipse has axes a and b. (a> b).


David

if you find it difficult to find the foci then go along these simple steps...



love arun

if you want to solve the problem algebraically,then you can go further ahead as follows....

1> find the equation of the line normal to the ellipse and passing through focus(3,0) or (-3,0) as you may wish..

2> find the intersection point of the line with the ellipse...

3> use distance formula to determine the radius..

love arun

Arun, if you do that you then have no guarantee that the circle lies wholly within the ellipse - it is still necessary to prove that separately.

David

David,

Did you get the answer b² /a (a,b semimajor and semiminor axes resp.) by equating curvatures?

Kerwin

David,
i can't quite understand your dilemma...

i am trying to find a point on the ellipse such that the normal to ellipse passing through focus and that point is also normal to the circle...

for this i find the slope of the tangent to ellipse at point say (x1,y1) then slope of normal to ellipse to the ellipse will be negative reciprocal of that slope....
then i will use the condition that the line is passing through (3,0) and find (x1,y1) and voila that's it???

please do correct me if i am wrong anywhere??

love arun

Arun, that method is guaranteed to find a circle tangent to the ellipse and having the right centre; but what's to stop us getting something like this:



Remember that the question said:
"Using one focus as center, a circle is drawn that is tangent to the ellipse, with no part of the circle being outside the ellipse."

The circle in the image above is tangent to the ellipse, yet it does not lie wholly inside it.

In fact this can't happen - my circle isn't centred at a focus of the ellipse - but you have to prove that it won't happen. This is what is missing from your method.


By the way - Kerwin, I did find that result by equating curvatures, then proved it by parametrizing the ellipse to ensure that this circle really is inside the ellipse (and that no larger one is so).

David

David,
Now,i got your point!!

Well,
i just have an idea..thoughhhhhh...

focus of the ellipse is (ae,0)
this is our center and if i am not wrong then the radius of the circle we get must be (a-ae)

so we can write the equation of the circle as
(x-ae)² +y² =(a-ae)² .....(1)

we know the equation of the elipse is,
(x² /a² )+(y² /b² )=1....(2)
also b² =a² (1-e² )

if we solve (1) and (2) simultneously and if we get one and only one point as the intersection point then the circle must lie wholly inside the ellipse and if we get more than one point as the intersection point then the circle doesn't lie wholly inside the ellipse...

i am not sure how far i am correct???
please do correct me if i am wrong??

love arun

Well, we could have only one point of intersection if the circle lay wholly outside the ellipse (except the one point of tangency). For example the circle with radius a+ae and centre (ae,0) which is tangent at the other end. However this clearly won't happen in this case (the circle you mention clearly includes at least some points inside the ellipse).

David

do you mean the proof is incomplete???

love arun

What I mean is that the proof is incomplete as it stands, but a minor extra statement - that the circle clearly does not lie entirely outside the ellipse - is all you need to make it so.

David

David,
Sorry to trouble you but i can't see what you are getting at???

please,could you explain it a bit??

love arun

Sorry, I'm not explaining myself well here, perhaps. In your message posted 6.58 Thursday, you suggest that we could prove that the circle centre (ae,0) and radius a-ae is the required circle by solving its equation simultaneously with that of the ellipse; if this gave only one solution, the circle must be wholly inside the ellipse, and is thus the one the original question asked for.

What I am trying to say is that this proof is incomplete as it stands: if we get only one solution for x when solving the equations simultaneously, it could mean that the circle lies within the ellipse, or it could mean it lies entirely outside it. However, the second case clearly won't occur for this circle, since it obviously doesn't lie entirely outside the ellipse. Hence, if we find only one solution for x, it does follow that the circle lies within the ellipse.

David

Oh!!yeah!!
thanks for the explanation,David....

BTW,in your earlier posts you have mentioned about the proof by triangle inequality...

could you show me the proof please???

love arun

Let P be a point on the ellipse, F₁, F₂ the foci, d the distance between them, and l the length of the major axis. Then if x=P F₁, we have:

x+d ³ P F₂ (as F₁ F₂ P is a triangle)

Þ x+d ³ l-x

Þ 2x ³ l-d

Þ x ³ (l-d)/2

Now (l-d)/2 is clearly the distance from the focus F₁ to the nearer endpoint of the major axis. So this is the smallest value of x, and it occurs only at that point (since we only obtain equality in the tri. ineq. if the points are collinear).

Hence for any other point on the ellipse P F₁ is greater than this value, ie the circle we are discussing lies within the ellipse.

David

Wow!!
it looks pretty neat...

love arun