Answer to Question #293619 in Analytic Geometry for Elley

Question

Show that the perpendiculars drawn from the origin to tangent planes to the cone 𝑥² − 𝑦² + 5z² + 4𝑥𝑦 = 0 lie on the cone 𝑥² − 𝑦² + 𝑧² + 4𝑥𝑦 = 0. 

Answer 1

According to the definition of reciprocal cone, we want to find the reciprocal of

$𝑥^2 − 𝑦^2 + 5z^2 + 4𝑥𝑦 = 0$

Equation of reciprocal of $𝑥^2 − 𝑦^2 + 5z^2 + 4𝑥𝑦 = 0$ is;

$Ax^2+By^2+Cz^2+2Fyz+2Gxz+2Hxy=0$

where,

$A=bc-f^2,\ B=ac-g^2,\ C=ab-h^2\\ F=gh-af,\ G=hf-bg,\ H=fg-ch$

Here, $a=1,\ b=-1,\ c=5,\ f=0,\ g=0,\ h=2$.

So,

$A=bc-f^2=-5-0=-5,\ B=ac-g^2=5-0=5,\ C=ab-h^2=-1-4=-5\\ F=gh-af=0-0=0,\ G=hf-bg=0-0=0,\ H=fg-ch=0-10=-10$

Thus, equation of reciprocal of $𝑥^2 − 𝑦^2 + 5z^2 + 4𝑥𝑦 = 0$ is;

$-5x^2+5y^2-5z^2+(2\times0\times yz)+(2\times0\times xz)-20xy=0\\ \Rightarrow -5x^2+5y^2-5z^2-20xy=0\\$

Dividing through by -5;

$\displaystyle \Rightarrow x^2-y^2+z^2+4xy=0,\ \text{which is the other cone given in the question.}$

Showing that that the perpendiculars drawn from the origin to tangent planes to the cone

$𝑥^2 − 𝑦^2 + 5z^2 + 4𝑥𝑦 = 0$ lie on the cone $𝑥^2 − 𝑦^2 + 𝑧^2 + 4𝑥𝑦 = 0$ .