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Cylinder with generators parallel to a line

Short answer

Generators direction (1,-2,3). Using condition that generator through (x,y,z) meets z=3 on x^2+2y^2=1 gives (3x-z+3)^2+2(3y+2z-6)^2=9.

Find the equation of the cylinder whose generators are parallel to the line x = −y/2 = z/3 and whose guiding curve is x²+2y² = 1, z = 3.

Given. Generators parallel to x = −y/2 = z/3, i.e. x/1 = y/(−2) = z/3, so direction ratios (1, −2, 3). Guiding curve: x² + 2y² = 1, z = 3.

Generator through P(x, y, z)

The line through P(x₁, y₁, z₁) with direction (1, −2, 3) has points (x₁ + k, y₁ − 2k, z₁ + 3k). It meets the plane z = 3 when z₁ + 3k = 3, i.e. k = (3 − z₁)/3.

The meeting point is

X = x₁ + (3 − z₁)/3 = (3x₁ − z₁ + 3)/3
Y = y₁ − 2(3 − z₁)/3 = (3y₁ + 2z₁ − 6)/3

On the guiding curve

X² + 2Y² = 1:

(3x₁ − z₁ + 3)²/9 + 2(3y₁ + 2z₁ − 6)²/9 = 1
(3x₁ − z₁ + 3)² + 2(3y₁ + 2z₁ − 6)² = 9

Locus

Dropping the subscripts:

(3x − z + 3)² + 2(3y + 2z − 6)² = 9

Expanded: 9x² + 18y² + 9z² − 6xz + 24yz + 18x − 72y − 54z + 72 = 0.

Check: at z = 3 this gives 9x² + 18y² − 9 = 0 after simplification, i.e. x² + 2y² = 1 ✓.

✓ Answer checked by Claude Opus 5.5 (high effort)

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