Question

Show that the plane 3x+12-6z = 17 touches the hyperboioid 3x2-6y2+9z2+17 = 0, and find the point of contact.

Answer 1

Let the plane , 3x + 12y - 6z = 17---------(1)
touches the hyperboiod, 3x² - 6y² + 9z² + 17 = 0

we know, A important application that ,
if ax² + by² + cz² + d = 0 is a curve and we have to find the equation of tangent at (x1,y1,z1) then, just write it axx1 + byy1 + czz1 + d = 0 it is equation of tangent .

now, the equation of tangent plane at (x1,y1,z1) to hyperboiod , 3x² - 6y² + 9z² + 17 = 0 is
3xx1 - 6yy1 + 9zz1 + 17 = 0 --------(2)

compare both equations (1) and (2),
3/3x1 = 12/-6y1 = -6/9z1 = -17/17 = -1
x1 = -1 , y = 2 , z = 2/3

now, the plane 3x + 12y - 6z = 17, touches the hyperboiod, 3x² - 6y² + 9z² + 17 = 0 if point of contact is point (-1,2,2/3) .
e.g., point (-1,2,2/3) lies on hyperboiod.
now, put the point in hyperboiod ,
e.g., 3(-1)² -6(2)² + 9(2/3)² + 17
= 3 - 24 + 4 + 17
= 0 , hence, it is true that given plane touches the given hyperboiod .

and the point of contact is (-1,2,2/3)