Apex of a tilted right circular cone

Question

A tilted right circular cone with apex angle $2 \theta_c = 45^\circ$ (where $\theta_c$ is the semi-vertical angle), has its apex at the point $A(a_x, a_y, a_z)$ with $a_x, a_y, a_z \gt 0$, such that its intersection with the $xy$ plane is given by

$$ 5 x^2 - 4 xy + 9 y^2 = 64 $$

Find the coordinates of the apex $A$.

My attempt:

First I identified the major and minor axes of the ellipse of intersection, their direction, and the eccentricity. From the eccentricity and the known semi-vertical angle, I calculated the angle between the normal to the $xy$ plane and the axis of the cone. Then used the angle bisector theorem to deduce the location of the point where the axis intersects the $xy$ plane and the distance between the apex and that intersection point. From there, I computed the coordinates of the apex.

Answer 1

Look at $$5x^2 - 4xy + 9y^2 +z(Ax +B y + C z + D) - 64=0$$ which has the required intersection with $z=0.$

Then we have the condition $576A^2+256AB+320B^2-41D^2-10496C=0$ for it to be a cone (the determinant of the $4\times 4$ matrix).

$$5x^2 - 4xy + 9y^2 -64\\-(41D^2-320B^2-256AB-576A^2)\,z^2/10496 +z\,(Ax +B y + D)=0$$

We find the apex, the singular point, by solving the systems of partials: $$(x,y,z)=(-(128B+576A)/(41D), -(320B+128A)/(41D), 128/D)\tag1$$ and check that it indeed lies on the surface.

It remains to make it a right circular cone with aperture $\frac{\pi}{8}.$

Shifting the apex to the origin by the obvious translation

$$(41D^2-320B^2-256AB-576A^2)z^2-10496Byz-10496Axz\\-94464y^2+41984xy-52480x^2=0\tag2$$

we see that we have a homogeneous form.

Remember $\frac{\sqrt{2+\sqrt2}}{2}=\cos(\frac{\pi}{8})=\frac{n_x x+ n_y y+n_z z}{\sqrt{x^2+y^2+z^2}}$ where $\hat{n}=(n_x,n_y,n_z)$ is a unit vector along the axis. We get $$\frac{2+\sqrt2}{4}(x^2+y^2+z^2)=(n_x x+ n_y y+n_z z)^2\tag3$$

Comparing coefficients in $(2)$ and $(3)$, where $k$ is a scaling factor, we get the following system, solving it in maxima CAS, precomputed a grobner basis in M2, all the while remembering that $$n_x^2+n_y^2+n_z^2=1$$

s:sqrt(2);
solve([n1*B+4*n3, A^2-2*A*B-B^2, 83968*k*A+s*B+209920*k*B+2*B, n2*A+4*n3, n1*A-n2*B+8*n3,2*s*A-9*s*B-1721344*k*B+4*A-18*B, 7774978036465664*k^2-19040*A*B-6272*B^2+287*D^2+430336*s+252941172736*k+1674112, 9033613312*s*k+2720*A*B+896*B^2-41*D^2-18067226624*k-146944, n2*n3-5248*k*B, 16*n1*n3+s*B+209920*k*B+2*B,4*n2^2-s-377856*k-2, n1*n2+20992*k, s*n2+83968*n1*k+209920*n2*k+2*n2, 4*n1^2-s-209920*k-2,s*n1+377856*n1*k+83968*n2*k+2*n1, s^2-2, 41*n2*D^2+3584*n3*A+3712*n3*B+20992*n1+52480*n2,41*n1*D^2+10880*n3*A+3584*n3*B+94464*n1+20992*n2,7*s*D^2+3253760*k*D^2+20992*n3^2+14*D^2+5248*s+3084648448*k+10496, 1312*B^3-85*A*D^2+198*B*D^2-94464*A+209920*B,1312*A*B^2+28*A*D^2-85*B*D^2+20992*A-94464*B, 146944*k*B^2-6724*k*D^2-340*n3^2+41*s+82, n2*B^2+4*n3*A-8*n3*B,7*s*B^2+67240*k*D^2+2952*n3^2+14*B^2-410*s-820, 2720*n3*A*B+896*n3*B^2-41*n3*D^2+5248*n2*B-94464*n3,20992*n2*k*B-s*n3-377856*n3*k-2*n3, 7904411648*n3^2*k+2205472*k*D^2+203360*n3^2-203*A*B+196*B^2-13448*s-26896,656*s*n3^2-1312*n3^2+5*A*B-2*B^2, 2720*A*B*D^2+896*B^2*D^2-41*D^4+3022848*A*B+671744*B^2-146944*D^2-110166016,6724*k*B*D^2+112*n3^2*A+116*n3^2*B-41*s*B-82*B,4299999936512*n3^4+47812050944*n3^2*B^2-988051456*n3^2*D^2-14193536*B^2*D^2+93275*D^4+30371007954944*k*D^2-740756291584*n3^2+1156743168*A*B-13918535680*B^2+334297600*D^2-185189072896*s-119750459392,1000195*k*D^4+257152*n3^2*B^2+56063*n3^2*D^2+3946396288*k*D^2+78609792*n3^2-2205472*s+2687499960320*k-4410944],[n1,n2,n3,A,B,D,k]);

And one of the solutions to this gives the positive octant apex:

n1:sqrt(13*sqrt(2)+16)/sqrt(82);
n2:((sqrt(2)-1)*sqrt(13*sqrt(2)+16))/sqrt(82);n3:sqrt(1-n1^2-n2^2);%r8:n3;
A:(sqrt(13*sqrt(2)+16)*(3*2^(3/2)-20)*sqrt(82)*%r8)/41;
B:((sqrt(2)-1)*sqrt(13*sqrt(2)+16)*(3*2^(3/2)-20)*sqrt(82)*%r8)/41;
D:(2^(5/2)*sqrt((2*(sqrt(2)-1)^2*(13*sqrt(2)+16)*(85*sqrt(2)+113)*(3*2^(3/2)-20)^2*%r8^2)/41-41*2^(9/2)-2296))/sqrt(41);
k:-(3*sqrt(2)+10)/1721344;

the above solution substituted into $(1)$ gives the apex

$$({{\sqrt{ 13\,\sqrt{2}+16}\,\sqrt{35-5\,2\sqrt2}\,\left(1856-2^5\sqrt2\right)\,\sqrt{82}}\over{1681\,\sqrt{2}\,\sqrt{2344- 23\,2^5\sqrt2}}},{{\sqrt{13\,\sqrt{2}+16}\,\sqrt{35-5\,2\sqrt2}\,\left(59\,2^5\sqrt2-1920\right)\,\sqrt{82} }\over{1681\,\sqrt{2}\,\sqrt{2344-23\,2^5\sqrt2}}},{{2^4\sqrt{82}}\over{\sqrt{2344-23\,2^5\sqrt2}}})$$ $$\approx(5.116868190800648,2.119476201505111,4.013576710399762).$$

Answer 2

First, I'll identify the given ellipse of intersection between the cone and the $xy$ plane. Define $r = [x, y]$ then the ellipse equation is

$r^T Q r = 1 $

where $Q = \dfrac{1}{64} \begin{bmatrix} 5 && - 2 \\ -2 && 9 \end{bmatrix} $

Diagonalize $Q$ into $Q = R D R^T $, then the angle of rotation of the axes is

$ \theta_0 = \dfrac{1}{2} \tan^{-1} \left( \dfrac{-4}{5 -9 } \right) = \dfrac{\pi}{8} $

The diagonal elements of $D$ are

$D_{11} = \frac{1}{2} \left( Q_{11} + Q_{22} + (Q_{11} - Q_{22} ) \cos(2 \theta_0) + 2 Q_{12} \sin(2 \theta_0) \right) = \dfrac{ 7 - 2 \sqrt{2} }{64} $

$D_{22} = \dfrac{1}{2} \left( Q_{11} + Q_{22} - ( Q_{11} - Q_{22} ) \cos(2 \theta_0) - 2 Q_{12} \sin(2 \theta_0) \right) = \dfrac{ 7 + 2 \sqrt{2} }{64 } $

From here, the semi-major and semi-minor axes are

$ a = \dfrac{1}{\sqrt{D_{11}}} \approx 3.91688 $

$ b = \dfrac{1}{\sqrt{D_{22}}} \approx 2.55181 $

The eccentricity is

$ e = \displaystyle \sqrt{1 - \left( \dfrac{b}{a} \right)^2 } \approx 0.75865 $

The eccentricity of the conic section and the angles $\theta_c$ (the semi-vertical angle) and $\phi$ (the angle between the normal to the section plane and the axis of the cone) is given by this important relation

$ e = \dfrac{ \sin(\phi) }{\cos(\theta_c) }$

Using this relation, we can compute $\phi$ readily. It comes to

$ \phi = \sin^{-1} \left( e \cos(\theta_c) \right) \approx 0.77666 $

Now, if we draw a section of the ellipse in the plane $y = \tan(\theta_0) x $, it will be the triangle shown in the diagram below.

where $\psi = \dfrac{\pi}{2} - \phi = 0.79414 $

Applying the law of sines on the triangles on both sides of the angle bisector (with is the axis of the cone, and is of length $z_0$) we get

$ \dfrac{ a - u }{z_0} = \dfrac{\sin(\theta_c) }{\sin(\theta_c + \psi) } $

and

$\dfrac{a + u}{z_0} = \dfrac{\sin(\theta_c)}{\sin(\psi - \theta_c} $

Dividing these two out, we obtain

$ \dfrac{ a - u }{a + u} = \dfrac{ \sin(\psi - \theta_c)}{\sin(\psi + \theta_c) } \approx 0.42143 $

Hence $ a - u = 0.42143 (a + u) $

and it follows that $ u = \dfrac{ (1 - 0.42143) a }{ 1.42143 } \approx 1.5943$

Finally from the first of the two equations above, we can calculate the length of the axis $z_0$

$ z_0 = (a - u) \dfrac{\sin(\psi + \theta_c)}{\sin(\theta_c)} \approx 5.62728 $

Now the coordinates of the apex are given by

$ A = P_0 + z_0 \hat{n} $

where $P_0 = u (\cos(\theta_0) , \sin(\theta_0) , 0 ) $

and $\hat{n} = ( \sin(\phi) \cos(\theta_0), \sin(\phi) \sin(\theta_0) , \cos(\phi) ) $

That is,

$ A = 1.5943 (\cos(\frac{\pi}{8}), \sin(\frac{\pi}{8}) , 0) \\ + 5.62728 ( \sin(0.77666) \cos(\frac{\pi}{8}) , \sin(0.77666) \sin(\frac{\pi}{8}) , \cos(0.77666) ) $

And this evaluates to

$ A \approx ( 5.116876, 2.119479, 4.013705 ) $

I re-did the calculations in an Excel Spreadsheet, keeping all decimals, and the exact coordinates I got were

$A \approx( 5.116868191, 2.119476202, 4.01357671 )$