Find the points at which line pierces the ellipsoid?

Question

The equation of the line is (x-6)/3 = (y+2)/-6 = (z-2)/4. The equation of the ellipsoid is x^2/81+ y^2/36 + z^2/9 = 1.

Answer 1

The equation of the line is:
t = (x - 6)/3 = (y + 2)/-6 = (z - 2)/4

Rewriting the equation of the line in parametric form we get:
L(t):
x = 6 + 3t
y = -2 - 6t
z = 2 + 4t

Plug these values into the equation of the ellipsoid.

x²/81 + y²/36 + z²/9 = 1
(6 + 3t)²/81 + (-2 - 6t)²/36 + (2 + 4t)²/9 = 1
4(6 + 3t)² + 9(-2 - 6t)² + 36(2 + 4t)² = 324

4(36 + 36t + 9t²) + 9(4 + 24t + 36t²)
+ 36(4 + 16t + 16t²) = 324

(144 + 144t + 36t²) + (36 + 216t + 324t²)
+ (144 + 576t + 576t²) = 324

936t² + 936t + 324 = 324
936t² + 936t = 0
936t(t + 1) = 0
t = 0, -1

The points at which the line pierces the ellipsoid are:

L(0):
x = 6
y = -2
z = 2
P(6, -2, 2)

L(-1):
x = 6 - 3 = 3
y = -2 + 6 = 4
z = 2 - 4 = -2
Q(3, 4, -2)

The points are P(6, -2, 2) and Q(3, 4, -2).