Answer to Question #114933 in Analytic Geometry for ANJU JAYACHANDRAN

Question #114933
Find the equation of the right circular cone whose vertex is (1,0,1), the axis is
x−1 = y−2 = z−3, and the semi-vertical angle is 30^◦. Also, find the section of
the cone by the coordinate planes.
1
Expert's answer
2020-05-18T11:00:26-0400

Direction ratio of generator is ((x-1),y,(z-1)).

Direction ratio of axis is (1,1,1).

Half angle is 30°.

cos(30°)=(a1a2+b1b2+c1c2)/(a12+b12+c12.a22+b22+c22)cos (30°)=(a_1a_2+b_1b_2+c_1c_2)/(\sqrt{a_1^{2}+b_1^{2}+c_1^{2}}.\sqrt{a_2^{2}+b_2^{2}+c_2^{2}})

3/2=(x+y+z2)/(3.(x1)2+y2+(z1)2)\sqrt{3}/2=(x+y+z-2)/(\sqrt{3}.\sqrt{(x-1)^{2}+y^{2}+(z-1)^{2}})

    9[(x1)2+y2+(z1)2]=4[(x+y+z2)2]\implies 9[(x-1)^{2}+y^2+(z-1)^{2}]=4[(x+y+z-2)^2]

    5x2+5y2+5z28xy8yz8xz2x2z+16y+2=0\implies 5x^2+5y^2+5z^2-8xy-8yz-8xz-2x-2z+16y+2=0

is the equation of the cone.


Section of the cone by x=0x=0 is 5y2+5z28yz2z+16y+2=05y^2+5z^2-8yz-2z+16y+2=0.

Section of the cone by y=0y=0 is 5x2+5z28xz2x2z+2=05x^2+5z^2-8xz-2x-2z+2=0 .

Section of the cone by z=0z=0 is 5x2+5y28xy2x+16y+2=05x^2+5y^2-8xy-2x+16y+2=0.





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