Question

Find the equation of the cylinder whose base is the circle x^2 + y^2=4 , x+y+2z=3

Answer 1

Given equation of base is

$x^2+y^2=4$

so the center of the cylinder lies at (0,0,0) and radius=2

Since Its equation of axis is fiven by

$x+y+2z=3\\\to\dfrac{x}{3}+\dfrac{y}{3}+\dfrac{2z}{3}=1$

therefore Dr's of the above line are (3,3,$\dfrac{3}{2})$

Let P(a,b,c) be the any point on the cylinder

Draw PM perpendicular to the axis of cylinder

Therefore PM=2

By distance formula,

$AP^2=(a-0)^2+(b-0)^2+(c-0)^2~~~~~~~~~~~`-(1)$

Let MA be the projection of AP on the axis

MA=$\dfrac{3(a-0)+3(b-0)+\dfrac{3}{2}(c-0)}{\sqrt{3^2+3^2+(\frac{3}{2})^2}}$

$\to MA=\dfrac{3a+3b+\dfrac{3}{2}c}{\sqrt{9+9+\frac{9}{4}}}$

$\to MA=\dfrac{\frac{3}{2}(2a+2b+c)}{\dfrac{9}{2}}$

$\to MA=\dfrac{1}{3}(2a+2b+c)$

As AP$^2-MA^2=(radius)^2$ ,

$a^2+b^2+c^2)-\dfrac{1}{9}(2a+2b+c)^2=2^2$

$\to 9a^2+9b^2+9c^2-4a^2-4b^2-c^2-8ab-4bc-4ac=36\\\to 5a^2+5b^2+8c^2-8ab-4bc-4ac=36$

Replacing a,b,c by x,y,z

Therefore the equation of the cylinder is

$5x^2+5y^2+8z^2-8xy-4yz-4xz=36.$