Projection Onto a Cone

Let's choose a reference frame for the cone, and place its origin at the apex of the cone, with the base at $z = -10$ , and we'll orient the $x$ and $y$ axes to be parallel to the sides of the inscribed square in the base. Projection of a side of the square onto the curved lateral surface of the cone is equivalent to finding the intersection between the plane parallel to the axis of the cone and passing through the side of the square. The intersection is clearly a hyperbola, it's eccentricity it given by the well-known formula, $e = \dfrac{\sin \beta} {\sin \alpha}$ , where $\beta$ is the angle between the plane and the horizontal and $\alpha$ is the angle between the cone's slant generator and the horizontal. Hence $\beta = \frac{\pi}{2}$ , and $\alpha = \tan^{-1} (10)$ . From this it follows that $\cos \alpha = \dfrac{1}{\sqrt{101}}$ , and that $\sin \alpha = \sqrt{ \dfrac{100}{101}}$ . Therefore, the eccentricity of the hyperbola is given by $e = \sqrt{\dfrac{101}{100}}$ . Now the equation of the hyperbola is

$\dfrac{z^2 }{a^2} - \dfrac{x^2}{b^2} = 1$

We know that $e^2 = 1 + (\dfrac{b}{a})^2$ , hence $\dfrac{b}{a} = \dfrac{1}{10}$ , and we also know that the point $( \dfrac{1}{\sqrt{2}}, -10 )$ is on the hyperbola, hence,

$\dfrac{100}{a^2} - \dfrac{1}{2 b^2} = 1$

but $a = 10 b$ , hence $\dfrac{100}{100 b^2} - \dfrac{1}{2 b^2} = 1$ , which implies that $b = \dfrac{1}{\sqrt{2}}$ and $a = \dfrac{10}{\sqrt{2}}$

Hence, the equation of the hyperbola is now fully determined, and it is,

$\frac{2}{100} z^2 - 2 x^2 = 1$

Since $z$ is negative, we have $z = - 10 \sqrt{ \frac{1}{2} + x^2 }$ .

To find the curve length, we find the derivative of $z$ with respect to $x$ .

$\dfrac{dz}{dx} = - \dfrac{10x}{\sqrt{\frac{1}{2}+ x^2} }$

And the curve length is given by

$L = \displaystyle \int_{x = -\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} \sqrt{1 + (\dfrac{dz}{dx})^2 }dx$

Which can be integrated numerically, and the required ratio will be $\dfrac{L}{\sqrt{2} }$ , and this comes to $4.328$ .