Question

Find the equation of the enveloping cylinder of the conicoid ax2 + by2 + cz2 = 1, whose generators are parallel to x = y = z.​

Answer 1

Answer:



Step-by-step explanation:

Answer 2

Answer:

The equation of the enveloping cylinder is:

$(x - \frac{1}{\sqrt{2} })^{2} +(y-\frac{1}{\sqrt{2} })^{2} +(z-\frac{1}{\sqrt{2} })^{2} =\frac{1}{2}$

Step-by-step explanation:

Given Data:
The equation of the conicoid is $ax^{2} +by^{2} +cz^{2} =1$

To Find:

The equation of the enveloping cylinder.

Calculations:

The equation of the conicoid is given by $ax^{2} +by^{2} +cz^{2} =1$.

The direction ratios of the generators are (1,1,1).  Let the cylinder's equation be$(x-a)^{2} +(y -b)^{2} +(z-c)^{2} =r^{2}$

The generator projections on the plane of the cylinder's base will be the conicoid's directions since the generators are parallel to the cylinder's axis.

The generators' projections on the xy plane are (1,1,0) and (-1,-1,0). The directrices are separated by 2a, thus we have:

Distance between the projections = 2a

$\sqrt{2}$ = 2a

a = $\frac{1}{\sqrt{2} }$

Similar to this, the distances between the directrices in the yz-plane and the zx-plane will be determined by the projections of the generators on these two planes.

So we have b =$\frac{1}{\sqrt{2} }$ and c = $\frac{1}{\sqrt{2} }$

As a result, the enveloping cylinder equation is:

$(x - \frac{1}{\sqrt{2} })^{2} +(y-\frac{1}{\sqrt{2} })^{2} +(z-\frac{1}{\sqrt{2} })^{2} =\frac{1}{2}$

This is the necessary equation for the conicoid's enveloping cylinder.

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