Formula for area of intersection of cone and plane, given their Cartesian equations.

Question

Prove that the area of the section of the cone $bc x^2+ca y^2+ab z^2=0$ obtained as the intersection with the plane $lx+my+nz=p$ is $$ \frac{\pi p^2 \sqrt{abc}}{{(al^2+bm^2+cn^2)}^{3/2}}.$$

Answer 1

We assume that $n\not=0$. By letting $z=(p-(lx+my))/n$ in $bc x^2+ca y^2+ab z^2=0$, we find the equation of the orthogonal projection on the $xy$ plane of the intersection curve: the conic $$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$ where $$A:= abl^2+cbn^2\;,\; B := 2ablm\;,\; C := abm^2+can^2\\ D := -2abpl\;,\; E := -2abpm\;,\; F := abp^2.$$ Now if this conic is an ellipse (that is when $4AC-B^2>0$), then the area of intersection of cone and plane is $$\mbox{area}=\int_{\Omega}\sqrt{1+\frac{l^2+m^2}{n^2}}dxdy=\frac{1}{|n|}\sqrt{l^2+m^2+n^2}\cdot|\Omega|.$$ where $\Omega$ is the inside of the ellipse. According to Calculating the length of the semi-major axis from the general equation of an ellipse , the area of this ellipse, which is $\pi$ times the product of the semi-major axis and the semi-minor axis, is $$|\Omega|= \frac{2\pi\left(\frac{AE^2{-}BDE{+}CD^2}{4AC{-}B^2}-F\right)}{\sqrt{4AC-B^2}}. $$ Now it turns out that $$\mbox{area}=\frac{\pi p^2 \sqrt{l^2+m^2+n^2}\sqrt{|abc|}}{ |al^2+bm^2+cn^2|^{3/2}}.$$

P.S. The above formula si "dimensionally correct". On the other hand, if in the formula stated in the question our formula is not correct. we double the coefficients of the plane then the numerator is multiplied by $2^2=4$ and the denominator by $(2^2)^{3/2}=8$.