Question

Find the equation of a cone reciprocal to the cone ax²+by²+cz²=0

Answer 1

Answer:

The equation of a cone reciprocal to the cone $ax^2+by^2+cz^2=0$ is given by $\frac{x^2}{a}+ \frac{y^2}{b}+ \frac{z^2}{c} = 0$.

Step-by-step explanation:

The reciprocal cone of a given cone is given by the locus of the normals through the vertex of the cone to the tangent planes of the given cone.

It is possible to prove that the reciprocal cone of a given cone $ax^2+by^2+cz^2+2dyz+2exz+2fxy=0$ is given by $Ax^2+By^2+Cz^2+2Dyz+2Exz+2Fxy=0$, where

$A = bc-d^2$

$B = ca-e^2$

$C = ab-f^2$

$D = ef-ad$

$E = df-be$

$F = de-cf$

In this case, we have $d = e = f = 0$. Therefore, the reciprocal cone is given by $bcx^2+cay^2+abz^2 = 0$. Dividing by $abc$, we get

$\frac{x^2}{a}+ \frac{y^2}{b}+ \frac{z^2}{c} = 0$.

Answer 2

Reciprocal cone formula:

We consider the equation of a cone in homogeneous of second degree, which is of the form

ax² + by² + cz² + 2fyz + 2gzx + 2hxy = 0

Another cone

Ax² + By² + Cz² + 2Fyz + 2Gzx + 2Hxy = 0

can be called the reciprocal cone to the first cone, when

A = bc - f² ,

B = ca - g² ,

C = ab - h² ,

F = gh - af ,

G = hf - bg and

H = fg - ch.

Solution:

The given cone is

ax² + by² + cz² = 0

Then the reciprocal cone must be determined using a, b, c in A, B, C, F, G, H (since f, g, h = 0 here).

A = bc - f² = bc ,

B = ca - g² = ca ,

C = ab - h² = ab and

F = G = H = 0

Hence, the reciprocal cone is

bcx² + cay² + abz² = 0

or, x²/a + y²/b + z²/c = 0 [ abc ≠ 0 ]