Question

X^2+y^2+z^2=9,x-y-z=3 with vertex(1,1,1) find the equation of cone and passing the curve

Answer 1

The equation of the surface S on which the base of the cone lies :
    S :  x² + y² + z² = 3²        --- (1)
It is a sphere symmetrical around origin of radius 3.

The equation of the plane P that cuts the sphere S giving the base of the cone:      P :  x - y - z = 3                ---(2)

The  vertex  V = (1, 1, 1).  Any line L that is on the lateral surface of the cone passes through V.  A generic equation for L is:
         
   $\frac{x-1}{l}=\frac{y-1}{m}=\frac{z-1}{n}=k \ \ \ ....(3)$

=> x = 1+ k l ,   y = 1+ k m  ,  z = 1 + k n     --- (4) 

The intersection of L with the Plane P is obtained by substituting (4) in (2):
        so   1+ kl - 1 - k m - 1 - k n = 3
$k=\frac{4}{l-m-n},\ \ \ \ ....(5)$

Intersection of Line L with the given surface S that marks the boundary of the base of the cone is obtained by substituting (4) in (1):

  (1 + k l)² + (1 + k m)² + (1 + k n)² = 3²
=>  k² (l² + m² + n²) + 2 k (l+m +n) = 6

Substitute value of k in the above to get:

  8 (l² + m²+ n²) + 4 ( l + m+ n) (l - m - n) = 3 ( l - m - n)²
  8 (l² + m²+ n²) + 4(l² - m² - n² - 2 mn)
               = 3 (l²+ m² + n²- 2 l m - 2 l n + 2 m n)
=>   9 l² + m² + n² -14 m n + 6 l m  + 6 l n = 0     ---(6)

Now substitute the value of l, m, and n from eq (3).  We get:

9 (x-1)² + (y-1)² + (z-1)² - 14 (y -1) (z-1) + 6 (x-1) (z-1)+ 6(x-1) (y-1) = 0

=> 9x² + y² +z²  - 8 z  - 22 y  - 30 x  + 37 + 6 x z + 6 x y = 0

This is the equation of the cone (lateral surface of the cone in 3-dimensions.

For a quick check, you can substitute V (1, 1, 1)  in the above. You can find it is satisfied.