Question

Find the equation to the cone with vertex is the origin and whose base curve is x^2+y^2+z^2+2ux+d=0​

Answer 1

Answer:

x^2+y^2+z^2+2ux+d=0 it's anger is over

Answer 2

The  given equation of the cone

$k^2(x^2+y^2+z^2)+2ukx(px+qy+rz)+d{(px+qy+rz)}^2=0\;$

Step by step solution:

Given:

$x^2+y^2+z^2+2ux+d=0\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots(1)$

$px+qy+rz=k\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots(2)$

To find:

The equation of homogeneous cone

Cones:

The required equation is the homogenous equation of second degree satisfied by the points common to the two equations.

                       ∴   $px+qy+rz=k$

                                  $\frac{px+qy+rz}k=1\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots\cdots(3)$

Now homogenising the equation $(1)$ by $(3)$ we have the homogeneous equation.

                   $x^2+y^2+z^2+2ux\left(\frac{px+qy+rz}k\right)+d\left(\frac{px+qy+rz}k\right)^2=0$

Thus equation to the homogeneous cone is

$k^2\left(x^2+y^2+z^2\right)+2ukx(px+qy+rz)+d{(px+qy+rz)}^2=0$