Question

Find the equation of the enveloping cylinder of the
sphere x2 + y2 + z2 - 2x + 4y - 1=0 and having
generators parallel to x = y = z

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Answer 1

Answer:

Y+Z=X is the correct and right answer

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Answer 2

Answer:

The equation of the enveloping cylinder of the sphere

$x^{2}+y^{2}+z^{2}-x y-y z-z x-4 x+5 y-z-2=0$.

Step-by-step explanation:

Given :

$x^{2} +y ^{2} +z^{2} -2 x+4 y-1=0$ and

$\frac{x}{1} =\frac{y}{1} =\frac{z}{1}$

To find the equation of the enveloping cylinder of the sphere  and having $x^{2} +y ^{2} +z^{2} -2 x+4 y-1=0$  generators parallel to x = y = z

Step 1

substitute all the values in the formula

$(x+u)^{2}+(y+v)^{2}+(z+w)^{2}$ $= \left[\{l(x-u)+m(y+v)+n(z+w)]^{2} / I^{2}+m^{2}+n^{2}\right]+a^{2}$

Here, $l=1$

$m=1$

and $n=1$

Equation of the sphere $x^{2}+y^{2}+z^{2}+2 u x+2 v y+2 \omega z+d=0$

Step 2

Here $u=-1,v=2,w=0,d=1$

Centre of sphere $:(1,-2,0)$

Radius of the sphere

$a=\sqrt{\left.u^{2}+v^{2}+w\right^{2}-d}$

$=\sqrt{1^2+\left(-2\right)^2+0^2+1}$

$=\sqrt{1+4+1+0}$

$=\sqrt{6}$

Step 3

The axis of a cylinder with direction ratios$(1,1,1)$ which passes through the centre $(1,-2,0)$ of the sphere. Let  $p(x,y,z)$ be any point of the cylinder.

Assume let N be the bottom of the perpendicular axis,

$P^{2}=O N^{2}+O N^{2}$

$(x-1)^{2}+(y+2)^{2}+(z)^{2}=\frac{[(x-1)+(y+2)+z]^{2}+6}{3}$

$(x-1)^{2}+(y+2)^{2}+(z)^{2}=\frac{[(x-1)+(y+2)+z]^{2}+18}{3}$

$3 x^{2}+3-6 x+3 y^{2}+12+12 y+3 z^{2}$$=x^{2}+1-2 x+y^{2}+4+4 y+z^{2} +2 x y+4 x-2 y-4+2 y z+4 z+2 z x-2 z+18$

Equating,

$3 x^{2}+3-6 x+3 y^{2}+12+12 y+3 z^{2}$$=x^{2}+1-2 x+y^{2}+4+4 y+z^{2} +2 x y+4 x-2 y-4+2 y z+4 z+2 z x-2 z+18$

⇒ $3 x^{2}+3-6 x+3 y^{2}+12+12 y+3 z^{2}$

$=x^{2}-2 x+y^{2}+4+4 y+z^{2} +2 x y+4 x-2 y-4+2 y z+4 z+2 z x-2 z+19$

⇒ $2 x^{2}+2 y^{2}+2 z^{2}-2 x y-2 y z-2 z x-8 x+10 y-2 z-4=0$

Step 4

Dividing the equation throughout by 2, we get

$x^{2}+y^{2}+z^{2}-x y-y z-z x-4 x+5 y-z-2=0$

Therefore,

The equation of the enveloping cylinder of the sphere

$x^{2}+y^{2}+z^{2}-x y-y z-z x-4 x+5 y-z-2=0$.