Question

Q.2
Find the equation of a cone reciprocal to the cone ax^2+by^2+cz^2=0.​

Answer 1

Answer:

$bcx^{2} +cay^{2} +abz^{2} =0$

Step-by-step explanation:

The given equation is $ax^{2} +by^{2} +cz^{2} =0$

The equation of the cone is

$[tex]ax^{2} +by^{2} +cz^{2} +2fyz+2gzx+2hxy=0$[/tex]

In the given equation

$a=a, b=b, c=c, f=0, g=0, h=0$

Reciprocal of the given cone is

$Ax^{2} +By^{2} +Cz^{2} +2Fyz+2Gzx+2Hxy=0\\A=bc-f^{2}=bc-0=bc , B=ca-g^{2}=ca-0=ca, C=ab-h^{2}=ab-0=ab,\\F=gh-af=0, G=hf-bg=0, H=fg-ch=0$

equation of the reciprocal cone is

$bcx^{2} +cay^{2} +abz^{2} =0$

Answer 2

Answer:

bcx^2+cay^2abz^2=0

Step-by-step explanation:

ax^2+by^2+cz^2= 0

given equation of the cone

{tex}ax^2+by^2+cz^2+2fyz+2gzx+2hxy=0[/tex}

Reciprocal

ax^2+by^2+cz^2+2fyz+2gzx+2hxy=0

A=bc-f^2=bc-0=bc

B=ca-g^2=ca-0=ca

C=ab-h^2=ab-0=ab

F=gh-af=0

G=hf-bg=0

H=fg-ch=0

hence

ax^2+by^2+cz^2=0.​