Find the equation of the right circular cone with
vertex at (1, 2, —1), axis as the x-axis and semi-
vertical angle π/3.
solution:
equation of rigth circular cone is c o s Θ = l ( x − α ) + m ( y − β ) + n ( z − Υ ) l 2 + m 2 + n 2 x 2 + y 2 + z 2 cos \Theta={ l(x-\alpha)+ m(y-\beta)+ n(z-\Upsilon) \over\sqrt{l^2+m^2+n^2}\sqrt{x^2+y^2+z^2}} cos Θ = l 2 + m 2 + n 2 x 2 + y 2 + z 2 l ( x − α ) + m ( y − β ) + n ( z − Υ )
the directional ratios (l,m,n)=(a,0,0) since the axis is the x-axis
θ = π 3 = 180 3 = 6 0 o \theta = {\pi \over 3}={180 \over3} = 60^o θ = 3 π = 3 180 = 6 0 o
∵ c o s 6 0 o = a ( x − 1 ) + 0 ( y − β ) + 0 ( z − γ ) x 2 + y 2 + z 2 a 2 + 0 2 + 0 2 \because cos 60^o={a(x-1)+0(y-\beta)+0(z-\gamma) \over \sqrt{x^2 +y^2 + z^2} \sqrt{a^2 + 0^2+0^2}} ∵ cos 6 0 o = x 2 + y 2 + z 2 a 2 + 0 2 + 0 2 a ( x − 1 ) + 0 ( y − β ) + 0 ( z − γ )
1 2 = a ( x − 1 ) a x 2 + y 2 + z 2 {1 \over 2} = {a(x-1) \over a\sqrt{x^2 +y^2 + z^2}} 2 1 = a x 2 + y 2 + z 2 a ( x − 1 )
1 2 = ( x − 1 ) x 2 + y 2 + z 2 {1 \over 2} = {(x-1) \over \sqrt{x^2 +y^2 + z^2}} 2 1 = x 2 + y 2 + z 2 ( x − 1 )
Square both sides
1 4 = ( x − 1 ) 2 x 2 + y 2 + z 2 {1 \over 4} = {(x-1)^2 \over x^2 +y^2 + z^2} 4 1 = x 2 + y 2 + z 2 ( x − 1 ) 2
x 2 + y 2 + z 2 = 4 ( x 2 − 2 x + 1 ) x^2 +y^2 + z^2 = 4(x^2 -2x +1) x 2 + y 2 + z 2 = 4 ( x 2 − 2 x + 1 )
x 2 + y 2 + z 2 = 4 x 2 − 8 x + 4 x^2 +y^2 + z^2 = 4x^2 -8x +4 x 2 + y 2 + z 2 = 4 x 2 − 8 x + 4
3 x 2 − y 2 − z 2 − 8 x + 4 = 0 3x^2 -y^2-z^2-8x +4=0 3 x 2 − y 2 − z 2 − 8 x + 4 = 0
the equation of the right circular cone is
3 x 2 − y 2 − z 2 − 8 x + 4 = 0 3x^2 -y^2-z^2-8x +4=0 3 x 2 − y 2 − z 2 − 8 x + 4 = 0
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