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    Mathematics
    Three Dimensional Geometry
    Cone
    Solved Problems on Cones with Guiding Curves and Sections

    Subject

    Mathematics
    Cone
    Solved Problems on Cones with Guiding Curves and Sections
    Active Unit
    Solved Problems on Enveloping Cones and Locus of Vertex
    Solved Problems on Common Vertex and Common Generators of Cones
    Solved Problems on Sections of Cones by Planes
    Solved Problems on Reciprocal Cones
    Solved Problems on Existence of Second Degree Cones
    Cone
    15 MIN READ ADVANCED

    Solved Problems on Cones with Guiding Curves and Sections

    Learning Objectives
    • • Master derivations of Solved Problems on Cones with Guiding Curves and Sections.
    • • Bridge theoretical limits with practice.
    example
    The section of a cone, whose guiding curve is the ellipse x2a2+y2b2=1,z=0,\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\quad z=0,a2x2​+b2y2​=1,z=0, by the plane x=0x=0x=0 is a rectangular hyperbola. It is required to show that the locus of the vertex of the cone is x2a2+y2+z2b2=1.\frac{x^2}{a^2}+\frac{y^2+z^2}{b^2}=1.a2x2​+b2y2+z2​=1.
    answer
    Let the vertex of the cone be denoted by (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ). A general point P(x,y,z)P(x,y,z)P(x,y,z) is taken on the cone. Let the generator through PPP meet the guiding ellipse at (x1,y1,0)(x_1,y_1,0)(x1​,y1​,0). Since this point lies on the ellipse, it satisfies x12a2+y12b2=1.(1)\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}=1. \quad (1)a2x12​​+b2y12​​=1.(1) The equations of the generator joining (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ) and (x1,y1,0)(x_1,y_1,0)(x1​,y1​,0) are written as x−αx1−α=y−βy1−β=z−γ−γ.\frac{x-\alpha}{x_1-\alpha} =\frac{y-\beta}{y_1-\beta} =\frac{z-\gamma}{-\gamma}.x1​−αx−α​=y1​−βy−β​=−γz−γ​. From these relations, one obtains x1=α−γx−αz−γ,y1=β−γy−βz−γ.x_1=\alpha-\gamma\frac{x-\alpha}{z-\gamma},\qquad y_1=\beta-\gamma\frac{y-\beta}{z-\gamma}.x1​=α−γz−γx−α​,y1​=β−γz−γy−β​. Substituting these expressions in equation (1), the equation of the cone is obtained as 1a2(αz−γxz−γ)2+1b2(βz−γyz−γ)2=1.\frac{1}{a^2}\left(\frac{\alpha z-\gamma x}{z-\gamma}\right)^2 +\frac{1}{b^2}\left(\frac{\beta z-\gamma y}{z-\gamma}\right)^2=1.a21​(z−γαz−γx​)2+b21​(z−γβz−γy​)2=1. The section of this cone by the plane x=0x=0x=0 is therefore α2z2a2+(βz−γy)2b2=(z−γ)2,x=0.\frac{\alpha^2 z^2}{a^2} +\frac{(\beta z-\gamma y)^2}{b^2} =(z-\gamma)^2,\quad x=0.a2α2z2​+b2(βz−γy)2​=(z−γ)2,x=0. This represents a rectangular hyperbola in the yzyzyz-plane. Hence, the sum of the coefficients of y2y^2y2 and z2z^2z2 must vanish. Therefore, γ2b2+α2a2+β2b2−1=0.\begin{aligned} \frac{\gamma^2}{b^2} +\frac{\alpha^2}{a^2} +\frac{\beta^2}{b^2} -1 &= 0. \end{aligned}b2γ2​+a2α2​+b2β2​−1​=0.​ Thus, α2a2+β2+γ2b2=1.\frac{\alpha^2}{a^2}+\frac{\beta^2+\gamma^2}{b^2}=1.a2α2​+b2β2+γ2​=1. Hence, the locus of the vertex (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ) is x2a2+y2+z2b2=1.\frac{x^2}{a^2}+\frac{y^2+z^2}{b^2}=1.a2x2​+b2y2+z2​=1. \quad ■\blacksquare■
    example
    The section of a cone, whose guiding curve is the ellipse x2a2+y2b2=1,z=0,\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\quad z=0,a2x2​+b2y2​=1,z=0, by the plane x=0x=0x=0 is a rectangular hyperbola. It is required to show that the locus of the vertex of the cone is x2a2+y2+z2b2=1.\frac{x^2}{a^2}+\frac{y^2+z^2}{b^2}=1.a2x2​+b2y2+z2​=1.
    answer
    Let the vertex of the cone be denoted by (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ). A general point P(x,y,z)P(x,y,z)P(x,y,z) is taken on the cone. Let the generator through PPP meet the guiding ellipse at (x1,y1,0)(x_1,y_1,0)(x1​,y1​,0). Since this point lies on the ellipse, it satisfies x12a2+y12b2=1.(1)\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}=1. \quad (1)a2x12​​+b2y12​​=1.(1) The equations of the generator joining (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ) and (x1,y1,0)(x_1,y_1,0)(x1​,y1​,0) are written as x−αx1−α=y−βy1−β=z−γ−γ.\frac{x-\alpha}{x_1-\alpha} =\frac{y-\beta}{y_1-\beta} =\frac{z-\gamma}{-\gamma}.x1​−αx−α​=y1​−βy−β​=−γz−γ​. From these relations, one obtains x1=α−γx−αz−γ,y1=β−γy−βz−γ.x_1=\alpha-\gamma\frac{x-\alpha}{z-\gamma},\qquad y_1=\beta-\gamma\frac{y-\beta}{z-\gamma}.x1​=α−γz−γx−α​,y1​=β−γz−γy−β​. Substituting these expressions in equation (1), the equation of the cone is obtained as 1a2(αz−γxz−γ)2+1b2(βz−γyz−γ)2=1.\frac{1}{a^2}\left(\frac{\alpha z-\gamma x}{z-\gamma}\right)^2 +\frac{1}{b^2}\left(\frac{\beta z-\gamma y}{z-\gamma}\right)^2=1.a21​(z−γαz−γx​)2+b21​(z−γβz−γy​)2=1. The section of this cone by the plane x=0x=0x=0 is therefore α2z2a2+(βz−γy)2b2=(z−γ)2,x=0.\frac{\alpha^2 z^2}{a^2} +\frac{(\beta z-\gamma y)^2}{b^2} =(z-\gamma)^2,\quad x=0.a2α2z2​+b2(βz−γy)2​=(z−γ)2,x=0. This represents a rectangular hyperbola in the yzyzyz-plane. Hence, the sum of the coefficients of y2y^2y2 and z2z^2z2 must vanish. Therefore, γ2b2+α2a2+β2b2−1=0.\begin{aligned} \frac{\gamma^2}{b^2} +\frac{\alpha^2}{a^2} +\frac{\beta^2}{b^2} -1 &= 0. \end{aligned}b2γ2​+a2α2​+b2β2​−1​=0.​ Thus, α2a2+β2+γ2b2=1.\frac{\alpha^2}{a^2}+\frac{\beta^2+\gamma^2}{b^2}=1.a2α2​+b2β2+γ2​=1. Hence, the locus of the vertex (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ) is x2a2+y2+z2b2=1.\frac{x^2}{a^2}+\frac{y^2+z^2}{b^2}=1.a2x2​+b2y2+z2​=1. \quad ■\blacksquare■
    example
    Show that the general equation of a quadric cone containing all the three coordinate axes is fyz+gzx+hxy=0.fyz + gzx + hxy = 0.fyz+gzx+hxy=0.
    It is known that any two generators of a cone intersect at the vertex. Since all the three coordinate axes lie on the cone, the origin must be the vertex of the cone. Hence, the general equation of a quadric cone with vertex at the origin is taken as ax2+by2+cz2+2fyz+2gzx+2hxy=0.ax^2 + by^2 + cz^2 + 2fyz + 2gzx + 2hxy = 0.ax2+by2+cz2+2fyz+2gzx+2hxy=0. As the cone contains the xxx-axis, whose direction cosines are (1,0,0)(1,0,0)(1,0,0), these values are substituted in the above equation. Thus, a(1)2+b(0)2+c(0)2=0⇒  a=0.\begin{aligned} & a(1)^2 + b(0)^2 + c(0)^2 = 0 \\ \Rightarrow \; & a= 0. \end{aligned}⇒​a(1)2+b(0)2+c(0)2=0a=0.​ Similarly, since the yyy-axis and the zzz-axis also lie on the cone, it follows that b=0andc=0.b = 0 \quad \text{and} \quad c = 0.b=0andc=0. Therefore, the equation of the cone reduces to 2fyz+2gzx+2hxy=0,2fyz + 2gzx + 2hxy = 0,2fyz+2gzx+2hxy=0, or equivalently, fyz+gzx+hxy=0.fyz + gzx + hxy = 0.fyz+gzx+hxy=0. \quad ■\blacksquare■
    Next Unit Solved Problems on Enveloping Cones and Locus of Vertex

    Section

    Cone

    Chapter

    Three Dimensional Geometry