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.

#### 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 $OA = b^2/a$.

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

Now the length of major axis $2a = AB = BC \sin x \iff \frac{a}{BC} = \frac{\sin x}{2}$.

Since CO is an angle bisector, then $OA/OB = CA/CB = \cos x$

$$\frac{OA}{AB-OA} = \cos x \iff OA = \frac{AB \cos x}{1 + \cos x} = \frac{2a \cos x}{1 + \cos x}$$

Now, let D be the point where the smaller Dandelin Sphere touches the cutting plane.

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

D is the focus that's closest to A. Therefore $DA = a-c$ where $c = \sqrt{a^2-b^2}$ (using the standard ellipse notation). But the center of the Dandelin sphere is also the incenter of the triangle ABC, so DA is the same as inradius of ABC. Using the inradius formula, and since $\angle BAC = \pi/2$

$$DA = \frac{AC.AB}{AC+AB+BC} = \frac{BC \cos x . BC \sin x}{BC \cos x + BC \sin x + BC} = BC \frac{\cos x \sin x}{1+\sin x + \cos x}$$

So

$$a-c = BC \frac{\cos x \sin x}{1+\sin x + \cos x}$$

$$c/BC = a/BC - \frac{\cos x \sin x}{1+\sin x + \cos x} = \frac{\sin x}{2} - \frac{\cos x \sin x}{1+\sin x + \cos x} = \frac{\sin x}{2} . \frac{1 + \sin x - \cos x}{1 + \sin x + \cos x}$$

Thus $$\frac{c}{a} = \frac{1 + \sin x - \cos x}{1 + \sin x + \cos x}$$

Because $a^2 = b^2+c^2$,

$$\left( \frac{b}{a} \right)^2 = 1 - \left( \frac{c}{a} \right)^2 = 1 - \frac{(1 + \sin x - \cos x)^2}{(1 + \sin x + \cos x)^2} = \frac{4(1+\sin x)(\cos x)}{(1 + \sin x + \cos x)^2}$$

But

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

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

So

$$\frac{b^2}{a^2} = \frac{2 \cos x}{1 + \cos x} = \frac{OA}{a}$$

Which means $OA = b^2/a$

Alternative solution available here: Second Solution

## No comments:

## Post a Comment