Find the equation of the enveloping cylinder of the sphere $x^{2} + y^{2} + z^{2} - 2 x + 4 y = 1$ $x^2+y^2+z^2-2x+4y=1$ with its lines parallel to $\frac{x}{2} = \frac{y}{3}, z = 0$ $x/2=y/3,z=0$ ?

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Find the equation of the enveloping cylinder of the sphere $x^{2} + y^{2} + z^{2} - 2 x + 4 y = 1$ $x^2+y^2+z^2-2x+4y=1$ with its lines parallel to $\frac{x}{2} = \frac{y}{3}, z = 0$ $x/2=y/3,z=0$ ?

Find the equation of the enveloping cylinder of the sphere
$x^{2} + y^{2} + z^{2} - 2 x + 4 y = 1$ $x^2+y^2+z^2-2x+4y=1$
with its lines parallel to
$\frac{x}{2} = \frac{y}{3}, z = 0$ $x/2=y/3,z=0$

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Answer 1 · 2017-03-17T15:27:30

See below.

Explanation:

The directrix is given by

$L_{d} \to p = p_{0} + λ \to v$ $L_d-> p= p_0+lambda vec v$

with $p = (x, y, z)$ $p=(x,y,z)$ , $p_{0} = (0, 0, 0)$ $p_0=(0,0,0)$ and $\to v = (2, 3, 0)$ $vec v = (2,3,0)$

The sphere is given by

$S \to ∥ p_{s} - p_{1} ∥ = r$ $S->norm(p_s-p_1)=r$

where

$p_{s} = (x_{s}, y_{s}, z_{s})$ $p_s=(x_s,y_s,z_s)$
$p_{1} = (1, - 2, 0)$ $p_1=(1,-2,0)$
$r = \sqrt{5}$ $r=sqrt(5)$

Now considering

$L_{c} \to p = p_{s} + λ \to v$ $L_c->p=p_s+lambda vec v$ as a generic line pertaining to the cylindrical surface, substituting $p_{s} = p - λ \to v$ $p_s = p - lambda vec v$ into $S$ $S$ we have

${∥ ∥ p - p_{1} - λ \to v ∥ ∥}^{2} = r^{2}$ $norm(p-p_1-lambda vec v)^2=r^2$ or

${∥ p - p_{1} ∥}^{2} - 2 λ ⟨ p - p_{1}, \to v ⟩ + λ^{2} {∥ ∥ \to v ∥ ∥}^{2} = r^{2}$ $norm(p-p_1)^2-2lambda << p-p_1, vec v >> + lambda^2 norm(vec v)^2 = r^2$

solving for $λ$ $lambda$

$λ = \frac{2 ⟨ p - p_{1}, \to v ⟩ \pm \sqrt{{(2 ⟨ p - p_{1}, \to v ⟩)}^{2} - 4 ({∥ p - p_{1} ∥}^{2} - r^{2})}}{2 {∥ ∥ \to v ∥ ∥}^{2}}$ $lambda = (2<< p-p_1, vec v >>pm sqrt((2<< p-p_1, vec v >>)^2-4(norm(p-p_1)^2-r^2)))/(2 norm(vec v)^2)$

but $L_{c}$ $L_c$ is tangent to $S$ $S$ having only a common point so

${(2 ⟨ p - p_{1}, \to v ⟩)}^{2} - 4 {∥ ∥ \to v ∥ ∥}^{2} ({∥ p - p_{1} ∥}^{2} - r^{2}) = 0$ $(2<< p-p_1, vec v >>)^2-4norm(vec v)^2(norm(p-p_1)^2-r^2)=0$ or

${⟨ p - p_{1}, \to v ⟩}_{1}^{2} - {∥ ∥ \to v ∥ ∥}^{2} ({∥ p - p_{1} ∥}^{2} - r^{2}) = 0$ $<< p-p_1, vec v >>^2-norm(vec v)^2(norm(p-p_1)^2-r^2)=0$

This is the cylindrical surface equation. After substituting numeric values

$9 x^{2} + 4 y^{2} + 13 z^{2} + 28 y - 42 x - 12 x y = 16$ $9 x^2 + 4 y^2 + 13 z^2 +28 y - 42 x - 12 x y=16$

Attached a plot of the resulting cylindrical surface.

enter image source here

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Find the equation of the enveloping cylinder of the sphere $x^{2} + y^{2} + z^{2} - 2 x + 4 y = 1$ $x^2+y^2+z^2-2x+4y=1$ with its lines parallel to $\frac{x}{2} = \frac{y}{3}, z = 0$ $x/2=y/3,z=0$ ?