Solved Problems on Sections of Cones by Planes

Let $$\frac{x}{l}=\frac{y}{m}=\frac{z}{n}$$ be a generator along which the plane intersects the cone. Then $$ul+vm+wn=0 \quad (1)$$ and $$al^2+bm^2+cn^2=0. \quad (2)$$ Eliminating $n$ from (1) and (2), $$\begin{aligned} al^2+bm^2+c\left(-\frac{ul+vm}{w}\right)^2&=0 \\ \Rightarrow\; w^2(al^2+bm^2)+c(ul+vm)^2&=0. \end{aligned}$$ This may be written as $$\begin{aligned} (aw^2+cu^2)l^2 +2cuv\,lm +(bw^2+cv^2)m^2=0. \end{aligned}$$ Dividing throughout by $m^2$, one gets $$(aw^2+cu^2)\left(\frac{l}{m}\right)^2 +2cuv\left(\frac{l}{m}\right) +bw^2+cv^2=0. \quad (3)$$ Let $(l_1,m_1,n_1)$ and $(l_2,m_2,n_2)$ be the direction ratios of the two generators of intersection. Then $\frac{l_1}{m_1}$ and $\frac{l_2}{m_2}$ are the roots of (3). Hence, $$\frac{l_1l_2}{m_1m_2} = \frac{bw^2+cv^2}{aw^2+cu^2}. \quad (4)$$ Similarly, eliminating $l$, one obtains $$\frac{m_1m_2}{n_1n_2} = \frac{cu^2+aw^2}{bu^2+av^2}. \quad (5)$$ From (4) and (5), $$\frac{l_1l_2}{bw^2+cv^2} = \frac{m_1m_2}{aw^2+cu^2} = \frac{n_1n_2}{bu^2+av^2} =k \; (k\neq 0).$$ Therefore, $$l_1l_2+m_1m_2+n_1n_2 = k\{(b+c)u^2+(c+a)v^2+(a+b)w^2\}.$$ If the lines are perpendicular, then $$l_1l_2+m_1m_2+n_1n_2=0,$$ which gives $$(b+c)u^2+(c+a)v^2+(a+b)w^2=0.$$ If the lines are parallel, the roots of (3) are equal. Hence, $$\begin{aligned} 4c^2u^2v^2 &=4(aw^2+cu^2)(bw^2+cv^2) \\ \Rightarrow\; c^2u^2v^2 &=abw^4+acv^2w^2+bcu^2w^2+c^2u^2v^2, \end{aligned}$$ which reduces to $$w^2(bcu^2+cav^2+abw^2)=0.$$ For $w\neq 0$, this yields $$bcu^2+cav^2+abw^2=0.$$ \quad $\blacksquare$

Let $(\lambda_1,\mu_1,\nu_1)$ and $(\lambda_2,\mu_2,\nu_2)$ denote the direction ratios of the generators of the cone $$ax^2+by^2+cz^2+2fyz+2gzx+2hxy=0$$ cut by the plane $lx+my+nz=0$. Then, $$\frac{\lambda_1\lambda_2}{bm^2+cn^2-2fmn} =\frac{\mu_1\mu_2}{cl^2+an^2-2gnl} =\frac{\nu_1\nu_2}{am^2+bl^2-2hlm}.$$ If the sections are coincident, the same generators must also lie on the cone $$fyz+gzx+hxy=0.$$ Hence, $$\frac{\lambda_1\lambda_2}{-fmn} =\frac{\mu_1\mu_2}{-gnl} =\frac{\nu_1\nu_2}{-hlm},$$ and also, $$\frac{\lambda_1\lambda_2}{bn^2+cm^2} =\frac{\mu_1\mu_2}{cl^2+an^2} =\frac{\nu_1\nu_2}{am^2+bl^2}.$$ Comparing these ratios, one obtains $$\frac{bn^2+cm^2}{fmn} =\frac{cl^2+an^2}{gnl} =\frac{am^2+bl^2}{hlm},$$ which are the required conditions. $\blacksquare$

Let $l,m,n$ be the direction ratios of a generator of the cone $$yz+zx+xy=0$$ which lies in the plane $$ax+by+cz=0.$$ Since the generator lies in the plane, the direction ratios satisfy $$al+bm+cn=0. \quad (1)$$ Also, because the generator lies on the cone, it follows that $$mn+nl+lm=0. \quad (2)$$ From equation (1), $n$ is eliminated to obtain $$n=-\frac{al+bm}{c}.$$ Substituting this value of $n$ in equation (2), one obtains $$\begin{aligned} & m\left(-\frac{al+bm}{c}\right) + l\left(-\frac{al+bm}{c}\right) + lm = 0 \\ \Rightarrow \; & al^2+(a+b-c)lm+bm^2 = 0. \end{aligned}$$ Dividing throughout by $m^2$, the equation reduces to $$a\left(\frac{l}{m}\right)^2+(a+b-c)\frac{l}{m}+b=0. \quad (3)$$ Let $(l_1,m_1,n_1)$ and $(l_2,m_2,n_2)$ be the direction ratios of the two generators in which the plane cuts the cone. Then $\frac{l_1}{m_1}$ and $\frac{l_2}{m_2}$ are the roots of equation (3). Hence, $$\frac{l_1l_2}{m_1m_2}=\frac{b}{a}. \quad (4)$$ Similarly, eliminating $l$ between equations (1) and (2), it is obtained that $$\frac{m_1m_2}{n_1n_2}=\frac{c}{a}. \quad (5)$$ From equations (4) and (5), it follows that $$\frac{l_1l_2}{1/a} =\frac{m_1m_2}{1/b} =\frac{n_1n_2}{1/c} = k \; (\neq 0),$$ for some constant $k$. Therefore, $$l_1l_2+m_1m_2+n_1n_2 = k\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right).$$ If the two generators are perpendicular, then $$l_1l_2+m_1m_2+n_1n_2=0.$$ Hence, $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0.$$ Thus, the plane cuts the cone in perpendicular straight lines under the given condition. \quad $\blacksquare$