An osculating circle of a point on a curve is defined as a circle that:

1. passes through that point

2. whose slope at that point is the same of the slope of the curve at that point

3. whose radius is the same as the radius of curvature of the curve at that point

In other words, it is a second-degree approximation circle of the curve at that point.

http://en.wikipedia.org/wiki/Osculating_circle

Given a cone whose half-angle is $\theta$, we take a cross section with a plane whose incident angle is $\theta$. That is, the plane is perpendicular to one of the cone rays. Naturally, the cross section forms an ellipse.

If O is the intersection of the main axis of the cone and the cross section, and A is the point on the ellipse's major axis that's closest to O, then prove that a circle with center O and radius OA is an osculating circle to the ellipse at A.

Hint: for people without any knowledge of calculus, the radius of osculating circle at A is $b^2/a$ where $b$ is half the length of minor axis and $a$ is half the length of major axis (standard ellipse notation).

The rest of the problem can be done without using calculus.

#### Second Solution

As given in the hint, the radius of the osculating circle is $b^2/a$. And clearly the circle in the problem passes through A and its tangent at A is perpendicular to the major axis, hence coincides with the ellipse's tangent. We are left to prove that $OA = b^2/a$. However, astute readers will note that $b^2/a$ is exactly the length of semi latus-rectum of the ellipse. So suppose $D$ is the focus that's closest to $A$, and $GG_1$ is the latus rectum passing through $D$, we will show that $DG = OA$.

Let $B$ be the point on the major axis that's farthest to $O$, and let $S$ be the vertex of the ellipse. Let $x = 2 \theta$ be the angle of the cone. We also note that $AO \perp AS$.

Let $\tau$ be the plane that passes through $GG_1$ (and hence through D) and perpendicular to the cone axis. Let $E$ be the point of intersection of the axis and this plane. The cross section of the cone with $\tau$ is a circle with center $E$. Let $F$ be the point on the circle such that $EF$ passes through $D$.

Since $SO$ is an angle bisector, we have $AO:OB = SA:SB = \cos x$, thus

$$AO = \frac{\cos x }{ \cos x+1} AB = \frac{\sin x \cos x }{ \cos x+1} SB$$

Now, $D$ is the point at which the smaller Dandelin Sphere touches the ellipse. If we consider the triangle $SAB$, then $D$ is where the incenter of that triangle touches $AB$. Since $SAB$ is a right angle at $A$, then $DA$ is the radius of that incenter.

$$DA = \frac{AS.AB}{AS+AB+SB} = \frac{\sin x \cos x}{1 + \sin x + \cos x} SB$$

Thus $AO : DA = (1+ \sin x + \cos x) : (1 + \cos x) = 1 + (\sin x) / (1 + \cos x)$

Which means $DO:AD:AO = \sin x : (1 + \cos x) : (1 + \sin x + \cos x)$

Now, looking back at the triangle $SAB$ and the plane that contains it,

$$DF = \frac{AD}{\cos \theta} = \frac{1 + \cos x}{(1 + \sin x + \cos x) \cos \theta} OA$$

and

$$DE = DO \cos \theta = \frac{\sin x \cos \theta}{1 + \sin x + \cos x} OA$$

So

$$DG^2 = EG^2 - DE^2 = EF^2 - DE^2 = (DE + DF)^2 - DE^2 = DF(DF + 2DE)$$

$$ = \frac{OA^2}{(1 + \sin x + \cos x)^2} \frac{1 + \cos x}{\cos \theta} \left(\frac{1 + \cos x}{\cos \theta} + 2\sin x \cos \theta \right)$$

Since $x = 2\theta$, then $1 + \cos x = 2 \cos^2 \theta$

$$(1 + \sin x + \cos x)^2 = (2 \cos^2 \theta + 2 \sin \theta \cos \theta)^2 = 4 \cos^2 \theta (\sin \theta + \cos \theta)^2$$

and

$$ \frac{1 + \cos x}{\cos \theta} + 2\sin x \cos \theta = 2 \cos \theta + 2 \sin x \cos \theta = 2 \cos \theta (1 + \sin x)$$

So

$$DG^2 = \frac{OA^2}{4 \cos^2 \theta (\sin \theta + \cos \theta)^2} \frac{2 \cos^2 \theta}{\cos \theta} 2 \cos \theta (1 + \sin x)$$

$$ = OA^2 \frac{1 + \sin x}{(\sin \theta + \cos \theta)^2}$$

But

$$1 + \sin x = 1 + 2 \sin \theta \cos \theta = \cos^2 \theta + \sin^2 \theta + 2 \sin \theta \cos \theta = (\sin \theta + \cos \theta)^2$$

So $DG^2 = OA^2$ which means $DG = OA$

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