Show that any plane whose normal lies on cone $(b+c)x^2+(c+a)y^2+(a+b)z^2=0$ cuts the surface $ax^2+by^2+cz^2=1$ is rectangular hyperbola

Question

Show that any plane whose normal lies on cone $(b+c)x^2+(c+a)y^2+(a+b)z^2=0$ cuts the surface $ax^2+by^2+cz^2=1$ is rectangular hyperbola

My attempt: let $\frac {x}{l} = \frac {y}{m} = \frac {z}{n}$ be normal of plane $lx+my+nz = 0$. Now the normal is generator of given cone. So we get condition $(b+c)l^2+(c+a)m^2+(a+b)n^2=0$ --->(1)

THe given plane cuts the conicoid $ax^2+by^2+cz^2=1$.

The intersecting conic :

Eliminating z between plane equation and conicoid equation

$z = \frac{-lx-my}{n} $ Subsittuting this in conicoid equation we get,

$ax^2+by^2+c\frac{(-lx-my)^2}{n^2} =1 $ --->(2)

This represents rectangular hyperbola if

coefficient of $x^2$ + coefficient of $y^2$ = 0

$(2)$ implies

$a+b+c\frac{l^2}{n^2}+c\frac{m^2}{n^2} =0 \\ \fbox {$n^2(a+b) + c(l^2+m^2) = 0$}$ for conic to be rectangular hyperbola

If i eliminate x, i get $\fbox {$l^2(b+c)+a(m^2+n^2)=0$}$ for conic to be rectangular hyperbola

If i eliminate y, i get $\fbox {$m^2(c+a)+b(n^2+l^2)=0$}$ for conic to be rectangular hyperbola

But what i have is equation (1) with me for conic to be rectangular hyperbola

How to proceed further?

Answer 1

First we will make some required transformations to assure that the formula $(b+c)x^2+(c+a)y^2+(a+b)z^2=0\;$ represents a real cone.

A real cone with circular base over the $x \times y$ plane and vertice at the origin should be represented as

$$ K\to m x^2+n y^2+p z^2 = 0 $$

with $m = n$ and $m > 0, n > 0, p < 0$

so we will choose $a,b,c$ such that

$$ b+c=m\\ a+c=n\\ a+b=p $$

in this representation a line $L$

$$ L\to P = \rho\vec v \to \left\{ \begin{array}{rcl} x = \rho \alpha\\ y = \rho \beta\\ z = \rho \gamma \end{array} \right.\;\;\;\; P = (x,y,z) $$

is such that $L \in K$ when

$$ m\alpha^2+n\beta^2+p\gamma^2=0 $$

Now with the help of $\vec v$ and $\vec w = (m\alpha,n\beta,p\gamma)$ Here $\left < \vec v, \vec w\right> = 0$, we can build a plane $\Pi$

$$ \Pi\to P = \lambda \vec w + \mu \left(\vec v \times \vec w\right) $$

as required. Given now the surface

$$ S\to ax^2+by^2+cy^2-1=0\equiv \frac{1}{2} x^2 (-m+n+p)+\frac{1}{2} y^2 (m-n+p)+\frac{1}{2} z^2 (m+n-p)-1 = 0 $$

the intersection $S\cap \Pi$ is obtained

$$ \left(S\circ\Pi\right)(\lambda,\mu) = \frac{1}{2} \left(-4 (m-n-p) (\alpha \lambda m+\beta \gamma \mu (n-p))^2+4 (m+n-p) (\alpha \beta \mu (m-n)+\gamma \lambda p)^2+4 (m-n+p) (\alpha \gamma \mu (p-m)+\beta \lambda n)^2-2\right)= 0 $$

but we can introduce a slight normalization to simplify the representation by considering

$$ m\alpha^2+n\beta^2+p\gamma^2=0\\ \alpha^2+\beta^2+\gamma^2 = 1\\ \beta = \beta_0 = 0\\ m = n\\ $$

and thus we have the intersection curve on $\Pi$ as

$$ 2np^2\lambda^2-2np^2\mu^2-1=0 $$

which is a rectangular hyperbola.

Answer 2

As an alternative solution,

Let $/$=$/$=$/$ be normal of plane ++=0. Now the normal is generator of given cone. So we get condition $(+)^2$+$(+)^2$+$(+)^2$=0 -----------(eq. 1)

Thus, our required plane is lx+my+nx=p whose normal is $/$=$/$=$/$

We now show that it's intersection with conicoid: $^2$+$^2$+$^2$=1 generates rectangular hyperbola, i.e. the asymptotes are mutually perpendicular.

Consider the asymptotes in the plane (lx+my+nx=p) as $(-x_0)/p$=$(-y_0)/q$=$(-z_0)/r$, having DCs <p,q,r> thus we should show that $p_1p_2$+$q_1q_2$+$r_1r_2$=0 then it will be a rectangular hyperbola.

Now, since the line (asymptote) with DCs <p,q,r> lie in plane lx+my+nz=p, thus lp+mq+nr=0 --------(eq.2)

For the line $(-x_0)/p$=$(-y_0)/q$=$(-z_0)/r$ = k(say), we have x=kp+$x_0$ ; y=qk+$y_0$ ; z=rk+$z_0$ and we put this in conicoid equation $^2$+$^2$+$^2$=1

we will get a quadratic in $k^2$, here we can put the coefficient of $k^2$ equal to zero, since the solution will be at infinity (for asymptote). Hence, we get another equation: $ap^2$+$bq^2$+$cr^2$=0 ---------(eq.3)

from equation 2 and equation 3, putting $p_1p_2$+$q_1q_2$+$r_1r_2$=0 (here, we just need to show sum of product of roots is zero) we will get our equation 1 i.e. $(+)^2$+$(+)^2$+$(+)^2$=0

Q.E.D.