show that the plane 3x+12y-6z=17 touches the hyperboloid 3x^2-6y^2+9z^2=-17 and find the point of contac
Answers
Easier method :
Given hyperboloid: x^2/3 - y^2/2 + z^2 /1 = - 17/9
Tangential plane at (a, b, c): a x/3 – 2 by/3 + c z/1 = -17/9
=> - 3a/17 x + 6b/17 y – 9c z /17 = 1
Given plane: 3 x/17 + 12 y/17 - 6 z/17 = 1
Compare coefficients : a = -1, b = 2, c = 2/3
Point of contact = (-1, 2, 2/3)
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Longer method by solving for intersection:
Equation of the plane: 3 x + 12 y - 6 z = 17
--- (1)
Hyperboloid = 3 x^2 - 6
y^2 + 9 z^2 = - 17
or, 12 x^2 - 24 y^2 + 36 z^2 = - 68
--- (2)
The points of contact form an ellipse.
6z = 3 x + 12 y - 17 from (1)
using (2),
12 x^2 - 24y^2 + 9x^2 + 144y^2 + 289+ 72xy - 408y -102 x = -68
simplify:
21 x^2 + 120 y^2 + 72 x y -102 x - 408 y + 357 =
0
7 x^2 + 40 y^2 + 24 x y - 34 x - 136 y + 119 = 0
7 x^2 + 2 (12 y – 17) x + (40y^2 – 136 y + 119) = 0
x = [ (17 – 12 y) + - sqrt { (12y-17)^2 – 280y^2 + 7*136y – 7*119 } ] / 7
x = [ (17 – 12 y) + - 2 sqrt(2*17)* sqrt { - (y - 2)^2 } ] /7
for x to be real, y = 2. Then x = - 1
From (1), z = (-3 + 24 – 17)/6 = 2/3
We have just one point of contact: (-1, 2, 2/3).