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    Mathematics
    Three Dimensional Geometry
    Cone
    Solved Problems on Common Vertex and Common Generators of Cones

    Subject

    Mathematics
    Cone
    Solved Problems on Cones with Guiding Curves and Sections
    Solved Problems on Enveloping Cones and Locus of Vertex
    Solved Problems on Common Vertex and Common Generators of Cones
    Active Unit
    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 Common Vertex and Common Generators of Cones

    Learning Objectives
    • • Master derivations of Solved Problems on Common Vertex and Common Generators of Cones.
    • • Bridge theoretical limits with practice.
    example
    Two cones pass through the curves y=0, z2=4axy=0,\, z^2=4axy=0,z2=4ax and x=0, z2=4byx=0,\, z^2=4byx=0,z2=4by with a common vertex. It is required to show that if their four common generators cut the plane z=0z=0z=0 in four concyclic points, then the vertex lies on z2 ⁣(xa+yb)=4(x2+y2).z^2\!\left(\frac{x}{a}+\frac{y}{b}\right)=4(x^2+y^2).z2(ax​+by​)=4(x2+y2).
    answer
    Let the common vertex be V(α,β,γ)V(\alpha,\beta,\gamma)V(α,β,γ). The cone through y=0, z2=4axy=0,\, z^2=4axy=0,z2=4ax is (γy−βz)2=4a(y−β)(αy−βx).\begin{equation} (\gamma y-\beta z)^2=4a(y-\beta)(\alpha y-\beta x). \end{equation}(γy−βz)2=4a(y−β)(αy−βx).​​ The cone through x=0, z2=4byx=0,\, z^2=4byx=0,z2=4by is (γx−αz)2=4b(x−α)(βx−αy).\begin{equation} (\gamma x-\alpha z)^2=4b(x-\alpha)(\beta x-\alpha y). \end{equation}(γx−αz)2=4b(x−α)(βx−αy).​​ Their sections by z=0z=0z=0 are respectively γ2y2=4a(y−β)(αy−βx),\begin{equation} \gamma^2y^2=4a(y-\beta)(\alpha y-\beta x), \end{equation}γ2y2=4a(y−β)(αy−βx),​​ γ2x2=4b(x−α)(βx−αy).\begin{equation} \gamma^2x^2=4b(x-\alpha)(\beta x-\alpha y). \end{equation}γ2x2=4b(x−α)(βx−αy).​​ Any curve passing through the common points of (3) and (4) is γ2y2−4a(y−β)(αy−βx)+λ{γ2x2−4b(x−α)(βx−αy)}=0.\gamma^2y^2-4a(y-\beta)(\alpha y-\beta x) +\lambda\{\gamma^2x^2-4b(x-\alpha)(\beta x-\alpha y)\}=0.γ2y2−4a(y−β)(αy−βx)+λ{γ2x2−4b(x−α)(βx−αy)}=0. Since this curve is a circle, the coefficients of x2x^2x2 and y2y^2y2 must be equal and that of xyxyxy must vanish. Hence, γ2−4aα=λ(γ2−4bβ),\gamma^2-4a\alpha=\lambda(\gamma^2-4b\beta),γ2−4aα=λ(γ2−4bβ), 4aβ+4bαλ=0.4a\beta+4b\alpha\lambda=0.4aβ+4bαλ=0. Eliminating λ\lambdaλ, one obtains γ2 ⁣(βb+αa)=4(α2+β2).\gamma^2\!\left(\frac{\beta}{b}+\frac{\alpha}{a}\right)=4(\alpha^2+\beta^2).γ2(bβ​+aα​)=4(α2+β2). Thus, the vertex lies on z2 ⁣(xa+yb)=4(x2+y2),z^2\!\left(\frac{x}{a}+\frac{y}{b}\right)=4(x^2+y^2),z2(ax​+by​)=4(x2+y2), as required. ■\blacksquare■
    example
    Two cones are constructed having the same vertex, with guiding curves zx=a2,  y=0andyz=b2,  x=0.zx=a^2,\; y=0 \quad \text{and} \quad yz=b^2,\; x=0.zx=a2,y=0andyz=b2,x=0. It is required to prove that if their four common generators intersect the plane z=0z=0z=0 in four concyclic points, then the vertex lies on the surface z(x2+y2)=a2x+b2y.z(x^2+y^2)=a^2x+b^2y.z(x2+y2)=a2x+b2y.
    answer
    Let the common vertex be (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ). Consider a point P(x,y,z)P(x,y,z)P(x,y,z) on the cone whose guiding curve is zx=a2,  y=0zx=a^2,\; y=0zx=a2,y=0. Suppose the generator through PPP meets the guiding curve at (x1,0,z1)(x_1,0,z_1)(x1​,0,z1​). Then z1x1=a2.(1)z_1x_1=a^2. \quad (1)z1​x1​=a2.(1) The equations of the generator through PPP are x−αx1−α=y−β−β=z−γz1−γ.\frac{x-\alpha}{x_1-\alpha} = \frac{y-\beta}{-\beta} = \frac{z-\gamma}{z_1-\gamma}.x1​−αx−α​=−βy−β​=z1​−γz−γ​. From these, x1=α−βx−αy−β=αy−βxy−β,z1=γ−βz−γy−β=γy−βzy−β.x_1=\alpha-\beta\frac{x-\alpha}{y-\beta} =\frac{\alpha y-\beta x}{y-\beta}, \quad z_1=\gamma-\beta\frac{z-\gamma}{y-\beta} =\frac{\gamma y-\beta z}{y-\beta}.x1​=α−βy−βx−α​=y−βαy−βx​,z1​=γ−βy−βz−γ​=y−βγy−βz​. Using equation (1), the equation of the first cone is obtained as γy−βzy−β⋅αy−βxy−β=a2⇒  (γy−βz)(αy−βx)=a2(y−β)2.(2)\begin{aligned} \frac{\gamma y-\beta z}{y-\beta}\cdot \frac{\alpha y-\beta x}{y-\beta} &=a^2 \\ \Rightarrow\; (\gamma y-\beta z)(\alpha y-\beta x) &=a^2(y-\beta)^2. \quad (2) \end{aligned}y−βγy−βz​⋅y−βαy−βx​⇒(γy−βz)(αy−βx)​=a2=a2(y−β)2.(2)​ Similarly, the equation of the second cone is (βx−αy)(γx−αz)=b2(x−α)2.(3)(\beta x-\alpha y)(\gamma x-\alpha z)=b^2(x-\alpha)^2. \quad (3)(βx−αy)(γx−αz)=b2(x−α)2.(3) The cone (2) cuts the plane z=0z=0z=0 in the conic γy(αy−βx)=a2(y−β)2,z=0,(4)\gamma y(\alpha y-\beta x)=a^2(y-\beta)^2, \quad z=0, \quad (4)γy(αy−βx)=a2(y−β)2,z=0,(4) and the cone (3) cuts the same plane in γx(βx−αy)=b2(x−α)2,z=0.(5)\gamma x(\beta x-\alpha y)=b^2(x-\alpha)^2, \quad z=0. \quad (5)γx(βx−αy)=b2(x−α)2,z=0.(5) These two conics intersect in four points. Any conic passing through these four points can be written as k1{γy(αy−βx)−a2(y−β)2}+k2{γx(βx−αy)−b2(x−α)2}=0,z=0.k_1\{\gamma y(\alpha y-\beta x)-a^2(y-\beta)^2\} + k_2\{\gamma x(\beta x-\alpha y)-b^2(x-\alpha)^2\} =0, \quad z=0.k1​{γy(αy−βx)−a2(y−β)2}+k2​{γx(βx−αy)−b2(x−α)2}=0,z=0. This conic is given to be a circle. Hence, the coefficients of x2x^2x2 and y2y^2y2 must be equal and the coefficient of xyxyxy must vanish. Therefore, k2(βγ−b2)=k1(αγ−a2),k_2(\beta\gamma-b^2)=k_1(\alpha\gamma-a^2),k2​(βγ−b2)=k1​(αγ−a2), and −k1βγ−k2αγ=0.-k_1\beta\gamma-k_2\alpha\gamma=0.−k1​βγ−k2​αγ=0. Eliminating k2k1\dfrac{k_2}{k_1}k1​k2​​, one obtains αγ−a2βγ−b2=−βα,\frac{\alpha\gamma-a^2}{\beta\gamma-b^2} = -\frac{\beta}{\alpha},βγ−b2αγ−a2​=−αβ​, which gives α2γ−a2α+β2γ−b2β=0,\alpha^2\gamma-a^2\alpha+\beta^2\gamma-b^2\beta=0,α2γ−a2α+β2γ−b2β=0, or γ(α2+β2)=a2α+b2β.\gamma(\alpha^2+\beta^2)=a^2\alpha+b^2\beta.γ(α2+β2)=a2α+b2β. Hence, the vertex (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ) lies on the surface z(x2+y2)=a2x+b2y.z(x^2+y^2)=a^2x+b^2y.z(x2+y2)=a2x+b2y. \quad ■\blacksquare■
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    Cone

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    Three Dimensional Geometry