# Apex of a tilted right circular cone

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.

• From Cone on wikipedia $$((x-a_x,y-a_y,z-a_z)\cdot(d_x,d_y,d_z))^2-(d_x,d_y,d_z)\cdot(d_x,d_y,d_z)(x-a_x,y-a_y,z-a_z)\cdot(x-a_x,y-a_y,z-a_z))/2=0$$ where $(d_x,d_y,d_z)$ is a vector parallel to the axis. Commented Apr 16, 2022 at 19:30
• Can you put that in an answer with numerical results, so that I verify my results? Commented Apr 16, 2022 at 21:28
• I used $\cos(\pi/4)^2$ above, where it should be $\cos(\pi/8)^2.$ Commented Apr 17, 2022 at 8:07
• Look at $$5x^2 - 4xy + 9y^2 +z(Ax +B y + C z + D) - 64=0$$ Commented Apr 17, 2022 at 9:08
• What do you mean by look at it ? What do I do with it ? Can you elaborate ? Commented Apr 17, 2022 at 9:54

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

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

• That's what I did in my solution. I looked at the last equation in your solution (with $z = 0$), and deduced $a_x, a_y, a_z$, which you used in your solution to find the equation of the cone. However, the question doesn't ask for the equation of the cone, just its apex coordinates. Commented Apr 17, 2022 at 8:37
• @GrabaCoffee in a comment you asked for the verification. Commented Apr 17, 2022 at 8:59
• What I asked for is verification of the values of $a_x, a_y, a_z$ using any other method. Commented Apr 17, 2022 at 9:01

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

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